Kinematics-2D
244 Questions
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Q226
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle is moving in a circle of radius $\frac{1}{4}$ m with a linear speed of $2 \, ms^{-1}$ . Calculate the angular speed of a particle, in $rads^{-1}$ .
Correct Answer: 8
Explanation:
The angular speed is
$\omega = \frac {v}{r} = \frac {2}{0 . 2 5} = 8 \mathrm{rads} ^ {- 1}$
$\omega = \frac {v}{r} = \frac {2}{0 . 2 5} = 8 \mathrm{rads} ^ {- 1}$
Q227
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A boy whirls a stone in a horizontal circle of radius 0.5 m and at height 20 m above level ground. The string breaks, and the stone flies off horizontally to strike the ground after travelling a horizontal distance of 10 m. Find the magnitude of the centripetal acceleration, in $ms^{-2}$ , of the stone while in circular motion. (Take $g = 10 \, ms^{-2}$ )
Correct Answer: 50
Explanation:
Since $y = u_{y}t + \frac{1}{2}a_{y}t^{2}$ , where $u_{y} = 0$ and $a_{y} = g$ , y = h.
So $h = \frac{1}{2}gt^{2}$ $\Rightarrow t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(20)}{10}} = 2s$ Further since $x = u_{x}t + \frac{1}{2}a_{x}t^{2}$ where $u_{x} = v$ (say), $a_{x} = 0$ , $x = 10 \, m$ , so $10 = vt = v(2)$ $\Rightarrow v = 5 \, ms^{-1}$ Since $a = \frac{v^{2}}{R}$ $\Rightarrow a = \frac{(5)^{2}}{0.5} = \frac{25}{0.5} = 50 \, ms^{-2}$ $\Rightarrow a = 50 \, ms^{-2}$
So $h = \frac{1}{2}gt^{2}$ $\Rightarrow t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2(20)}{10}} = 2s$ Further since $x = u_{x}t + \frac{1}{2}a_{x}t^{2}$ where $u_{x} = v$ (say), $a_{x} = 0$ , $x = 10 \, m$ , so $10 = vt = v(2)$ $\Rightarrow v = 5 \, ms^{-1}$ Since $a = \frac{v^{2}}{R}$ $\Rightarrow a = \frac{(5)^{2}}{0.5} = \frac{25}{0.5} = 50 \, ms^{-2}$ $\Rightarrow a = 50 \, ms^{-2}$
Q228
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A highway curve is designed such that the cars travelling at a constant speed of $25 \, ms^{-1}$ must not have an acceleration that exceeds $3 \, ms^{-2}$ . Determine the minimum radius of curvature, in metre, of the curve to the nearest integer.
Correct Answer: 208
Explanation:
Since the car is travelling with a constant speed, its tangential component of acceleration is zero, i.e., $a_{T}=0$ . Thus,
$a = a _ {N} = \frac {v ^ {2}}{r}$
$\Rightarrow 3 = \frac {2 5 ^ {2}}{r}$
$\Rightarrow r = 2 0 8 \mathrm{m}$
$a = a _ {N} = \frac {v ^ {2}}{r}$
$\Rightarrow 3 = \frac {2 5 ^ {2}}{r}$
$\Rightarrow r = 2 0 8 \mathrm{m}$
Q229
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
At a given instant, a car travels along a circular curved road with a speed of $20 \, ms^{-1}$ while decreasing its speed at the rate of $3 \, ms^{-2}$ . If the magnitude of the car's acceleration is $5 \, ms^{-2}$ , determine the radius of curvature of the road, in metre.
Correct Answer: 100
Explanation:
Here, the car's tangential component of acceleration of $a_{T} = -3 \, \mathrm{ms}^{-2}$ . Thus,
$a = \sqrt {a _ {T} ^ {2} + a _ {N} ^ {2}}$
$\Rightarrow 5 = \sqrt {(- 3) ^ {2} + a _ {N} ^ {2}}$
$\Rightarrow a _ {N} = 4 \mathrm{ms} ^ {- 2}$
Since $a_{N} = \frac{v^{2}}{r}$
$\Rightarrow 4 = \frac {2 0 ^ {2}}{r}$
$\Rightarrow r = 1 0 0 \mathrm{m}$
$a = \sqrt {a _ {T} ^ {2} + a _ {N} ^ {2}}$
$\Rightarrow 5 = \sqrt {(- 3) ^ {2} + a _ {N} ^ {2}}$
$\Rightarrow a _ {N} = 4 \mathrm{ms} ^ {- 2}$
Since $a_{N} = \frac{v^{2}}{r}$
$\Rightarrow 4 = \frac {2 0 ^ {2}}{r}$
$\Rightarrow r = 1 0 0 \mathrm{m}$
Q230
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Ball bearings of diameter 20 mm leave the horizontal with a velocity of magnitude u and fall through the 60 mm diameter hole at a depth of 800 mm as shown. Calculate the permissible range of u, in cms $^{-1}$ which will enable the ball bearings to enter the hole. Take the dotted positions to represent the limiting conditions. (Take $g = 10 \, ms^{-2}$ )
Correct Answer: Min. 25 & Max. 35
Explanation:
$t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2\times 0.8}{10}} = 0.4\mathrm{s}$
$u _ {\mathrm{MIN}} = \frac {(1 2 0 - 3 0 + 1 0) \times 1 0 ^ {- 3}}{0 . 4} = 0. 2 5 \mathrm{ms} ^ {- 1}$
and $u_{\mathrm{MAX}}=\frac{(120+30-10)\times10^{-3}}{0.4}=0.35\ \mathrm{ms}^{-1}$
$\Rightarrow u _ {M I N} = 2 5 \mathrm{cms} ^ {- 1} \mathrm{and} u _ {M A X} = 3 5 \mathrm{cms} ^ {- 1}$
$u _ {\mathrm{MIN}} = \frac {(1 2 0 - 3 0 + 1 0) \times 1 0 ^ {- 3}}{0 . 4} = 0. 2 5 \mathrm{ms} ^ {- 1}$
and $u_{\mathrm{MAX}}=\frac{(120+30-10)\times10^{-3}}{0.4}=0.35\ \mathrm{ms}^{-1}$
$\Rightarrow u _ {M I N} = 2 5 \mathrm{cms} ^ {- 1} \mathrm{and} u _ {M A X} = 3 5 \mathrm{cms} ^ {- 1}$
Q231
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A ball is launched by a man standing on the top of a building who holds the ball at a distance 1 m above the edge A. Calculate the minimum velocity u, in ms $^{-1}$ , along the horizontal such that the ball just clears the edge C. Also find x, in metre, where the ball strikes the ground. Take $g = 9.8 \, ms^{-2}$ .
Correct Answer: 6, 6
Explanation:
Let the ball clear the point C at time $t_{1}$ . Then
$\begin{array}{r l r} & y = u _ {y} t + \frac {1}{2} a _ {y} t ^ {2} \\ \Rightarrow & (1 + 3. 9) = 0 + \frac {1}{2} (9. 8) t _ {1} ^ {2} & \left\{\because u _ {y} = 0 \right\} \\ \Rightarrow & 4. 9 = \frac {1}{2} (9. 8) t _ {1} ^ {2} \\ \Rightarrow & t _ {1} = 1 \mathrm{s} \\ \text {Since} & B C = u _ {x} t \\ \Rightarrow & 6 = u (1) \\ \Rightarrow & u = 6 \mathrm{ms} ^ {- 1} \end{array}$
Let the ball hit the ground at D in time t. Then
$6 + x = 6 t\tag{... (1}$
$\text { Further } (1 + 3. 9 + 1 4. 7) = \frac {1}{2} (9. 8) t ^ {2}$
$\Rightarrow t = 2 \mathrm{s}$
$\Rightarrow 6 + x = 1 2$
$\Rightarrow x = 6 \mathrm{m}$
$\begin{array}{r l r} & y = u _ {y} t + \frac {1}{2} a _ {y} t ^ {2} \\ \Rightarrow & (1 + 3. 9) = 0 + \frac {1}{2} (9. 8) t _ {1} ^ {2} & \left\{\because u _ {y} = 0 \right\} \\ \Rightarrow & 4. 9 = \frac {1}{2} (9. 8) t _ {1} ^ {2} \\ \Rightarrow & t _ {1} = 1 \mathrm{s} \\ \text {Since} & B C = u _ {x} t \\ \Rightarrow & 6 = u (1) \\ \Rightarrow & u = 6 \mathrm{ms} ^ {- 1} \end{array}$
Let the ball hit the ground at D in time t. Then
$6 + x = 6 t\tag{... (1}$
$\text { Further } (1 + 3. 9 + 1 4. 7) = \frac {1}{2} (9. 8) t ^ {2}$
$\Rightarrow t = 2 \mathrm{s}$
$\Rightarrow 6 + x = 1 2$
$\Rightarrow x = 6 \mathrm{m}$
Q232
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle is projected from a point at the foot of a fixed plane, inclined at an angle of $45^{\circ}$ to the horizontal, in the vertical plane containing the line of greatest slope through the point. If the particle strikes the plane at right angles and $\phi (>45^{\circ})$ is the angle of launch measured to the horizontal, then find the value of $\tan\phi$ .
Correct Answer: 3
Explanation:
At time $t = T = \frac{2u \sin(\phi - 45^{\circ})}{g \cos 45^{\circ}}$
component of velocity along the plane is zero.
$\Rightarrow 0 = u \cos (\phi - 4 5 ^ {\circ}) - (g \sin 4 5 ^ {\circ}) t$
$\Rightarrow u \cos (\phi - 4 5 ^ {\circ}) = (g \sin 4 5 ^ {\circ}) \left(\frac {2 u \sin (\phi - 4 5 ^ {\circ})}{g \cos 4 5 ^ {\circ}}\right)$
$\Rightarrow 2\tan (\phi -45^{\circ}) = \cot 45^{\circ} = 1$
$\Rightarrow 2 \left(\frac {\tan \phi - \tan 4 5 ^ {\circ}}{1 + \tan \phi \tan 4 5 ^ {\circ}}\right) = 1$
$\Rightarrow \tan \phi = 3$
component of velocity along the plane is zero.
$\Rightarrow 0 = u \cos (\phi - 4 5 ^ {\circ}) - (g \sin 4 5 ^ {\circ}) t$
$\Rightarrow u \cos (\phi - 4 5 ^ {\circ}) = (g \sin 4 5 ^ {\circ}) \left(\frac {2 u \sin (\phi - 4 5 ^ {\circ})}{g \cos 4 5 ^ {\circ}}\right)$
$\Rightarrow 2\tan (\phi -45^{\circ}) = \cot 45^{\circ} = 1$
$\Rightarrow 2 \left(\frac {\tan \phi - \tan 4 5 ^ {\circ}}{1 + \tan \phi \tan 4 5 ^ {\circ}}\right) = 1$
$\Rightarrow \tan \phi = 3$
Q233
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
The minimum speed in $ms^{-1}$ with which a projectile must be thrown from origin at ground so that it is able to pass through a point $P(30\ m, 40\ m)$ is $(g = 10\ ms^{-2})$
Correct Answer: 30
Explanation:
As $y = x \tan \theta - \frac{gx^{2}}{2u^{2}}(1 + \tan^{2} \theta)$
for $(a, b)$ , i.e. when $x = a$ and $y = b$ , we have
$g a ^ {2} \tan^ {2} \theta - 2 a u ^ {2} \tan \theta + (g a ^ {2} + 2 b u ^ {2}) = 0$
This is a quadratic in $\tan\theta$ and since discriminant of a quadratic must be positive, so we have
$4 a ^ {2} u ^ {2} - 4 g a ^ {2} \left(g a ^ {2} + 2 b u ^ {2}\right) \geq 0$
Solving, we get
$u \geq \sqrt {b g + g \sqrt {a ^ {2} + b ^ {2}}}$
On substituting the values, we get
$u _ {\mathrm{min}} = 3 0 \mathrm{ms} ^ {- 1}$
for $(a, b)$ , i.e. when $x = a$ and $y = b$ , we have
$g a ^ {2} \tan^ {2} \theta - 2 a u ^ {2} \tan \theta + (g a ^ {2} + 2 b u ^ {2}) = 0$
This is a quadratic in $\tan\theta$ and since discriminant of a quadratic must be positive, so we have
$4 a ^ {2} u ^ {2} - 4 g a ^ {2} \left(g a ^ {2} + 2 b u ^ {2}\right) \geq 0$
Solving, we get
$u \geq \sqrt {b g + g \sqrt {a ^ {2} + b ^ {2}}}$
On substituting the values, we get
$u _ {\mathrm{min}} = 3 0 \mathrm{ms} ^ {- 1}$
Q234
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Two particles are simultaneously thrown from top of two towers as shown. Their velocities are $2 \, ms^{-1}$ and $14 \, ms^{-1}$ . Horizontal and vertical separation between these particles are 22 m and 9 m, respectively. Then the minimum separation between the particles in process of their motion in meters is $(g = 10 \, \text{ms}^{-2})$ .
Correct Answer: 6
Explanation:
$v_{x}=8\sqrt{2}$ ms $^{-1}$ is the relative velocity along x-axis
$\Rightarrow x = 2 2 - (8 \sqrt {2}) t$
$v_{y}=6\sqrt{2}$ ms $^{-1}$ is the relative velocity along y-axis
$\Rightarrow y = 9 - (6 \sqrt {2}) t$
$\Rightarrow \quad r ^ {2} = x ^ {2} + y ^ {2}$
For minimum r, we have $r^{2}$ to be minimum, so
$\frac {d}{d t} \big (r ^ {2} \big) = 0$
$\Rightarrow t = \frac {2 3}{1 0 \sqrt {2}} \mathrm{s}$
Substituting the value of t in equation (1), we get
$r _ {\mathrm{min}} = 6 \mathrm{m}$
$\Rightarrow x = 2 2 - (8 \sqrt {2}) t$
$v_{y}=6\sqrt{2}$ ms $^{-1}$ is the relative velocity along y-axis
$\Rightarrow y = 9 - (6 \sqrt {2}) t$
$\Rightarrow \quad r ^ {2} = x ^ {2} + y ^ {2}$
For minimum r, we have $r^{2}$ to be minimum, so
$\frac {d}{d t} \big (r ^ {2} \big) = 0$
$\Rightarrow t = \frac {2 3}{1 0 \sqrt {2}} \mathrm{s}$
Substituting the value of t in equation (1), we get
$r _ {\mathrm{min}} = 6 \mathrm{m}$
Q235
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A projectile is fixed at an angle $60^{\circ}$ with horizontal. Ratio of initial and final kinetic energy velocity vector of projectile makes an angle $15^{\circ}$ with velocity of projection is
Correct Answer: 2
Explanation:
$v=\frac{u\cos(60^{\circ})}{\cos(45^{\circ})}$
$\Rightarrow K _ {f} = K _ {i} \left(\frac {\cos^ {2} (6 0 ^ {\circ})}{\cos^ {2} (4 5 ^ {\circ})}\right)$
$\Rightarrow \frac {K _ {i}}{K _ {f}} = 2$
$\Rightarrow K _ {f} = K _ {i} \left(\frac {\cos^ {2} (6 0 ^ {\circ})}{\cos^ {2} (4 5 ^ {\circ})}\right)$
$\Rightarrow \frac {K _ {i}}{K _ {f}} = 2$
Q236
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle is projected from the bottom of an inclined plane of inclination $30^{\circ}$ . At what angle $\alpha$ (from the horizontal), in degree, should the particle be projected to get the maximum range on the inclined plane?
Correct Answer: 60
Explanation:
For Range to be maximum, the projectile must be launched at an angle
$\alpha = \frac {\pi}{4} + \frac {\beta}{2}$
$\Rightarrow \alpha = 4 5 ^ {\circ} + 1 5 ^ {\circ} = 6 0 ^ {\circ}$
$\alpha = \frac {\pi}{4} + \frac {\beta}{2}$
$\Rightarrow \alpha = 4 5 ^ {\circ} + 1 5 ^ {\circ} = 6 0 ^ {\circ}$
Q237
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A small body is released from point $A$ of smooth parabolic path $y = x^2$ , where $y$ is vertical axis and $x$ is horizontal axis at ground as shown. The body leaves the surface from point $B$ . If $g = 10 \, \text{ms}^{-2}$ , then total horizontal distance in meters travelled by body before it hits ground is ____
Correct Answer: 8
Explanation:
Since, $H_{1}=(2)^{2}=4\ m$ and
$H _ {2} = (1) ^ {2} = 1 \mathrm{m}$
By Conservation of Energy, we have
$\binom{\text {Loss in gravitational}}{\text {potential energy}} = \binom{\text {Gain in}}{\text {kinetic energy}}$
$\Rightarrow m g \left(H _ {1} - H _ {2}\right) = \frac {1}{2} m v ^ {2}$
$\begin{array}{r l} \Rightarrow & v = \sqrt {2 g (H _ {1} - H _ {2})} \\ \Rightarrow & v = \sqrt {2 \times g \times (H _ {2} - H _ {1})} \\ \Rightarrow & v = \sqrt {2 0 \times 3} \\ \Rightarrow & v = \sqrt {6 0} \end{array}$
Now $\tan \theta = \frac{dy}{dx}\bigg|_{x = 1} = 2x\big|_{x = 1} = 2$
$\Rightarrow \tan \theta = 2$
$\begin{array}{r l} & {\mathrm{Since,} y = x \tan \theta - \frac {g x ^ {2}}{2 v ^ {2} \cos^ {2} \theta}} \\ & {\Rightarrow - 1 = x (2) - \frac {1 0 x ^ {2}}{2 v ^ {2}} (1 + 4)} \\ & {\Rightarrow - 1 = 2 x - \frac {1 0 x ^ {2}}{2 (6 0)} (5)} \\ & {\Rightarrow - 1 = 2 x - \frac {5}{1 2} x ^ {2}} \\ & {\Rightarrow - 1 2 = 2 4 x - 5 x ^ {2}} \\ & {\Rightarrow 5 x ^ {2} - 2 4 x - 1 2 = 0} \end{array}$
$H _ {2} = (1) ^ {2} = 1 \mathrm{m}$
By Conservation of Energy, we have
$\binom{\text {Loss in gravitational}}{\text {potential energy}} = \binom{\text {Gain in}}{\text {kinetic energy}}$
$\Rightarrow m g \left(H _ {1} - H _ {2}\right) = \frac {1}{2} m v ^ {2}$
$\begin{array}{r l} \Rightarrow & v = \sqrt {2 g (H _ {1} - H _ {2})} \\ \Rightarrow & v = \sqrt {2 \times g \times (H _ {2} - H _ {1})} \\ \Rightarrow & v = \sqrt {2 0 \times 3} \\ \Rightarrow & v = \sqrt {6 0} \end{array}$
Now $\tan \theta = \frac{dy}{dx}\bigg|_{x = 1} = 2x\big|_{x = 1} = 2$
$\Rightarrow \tan \theta = 2$
$\begin{array}{r l} & {\mathrm{Since,} y = x \tan \theta - \frac {g x ^ {2}}{2 v ^ {2} \cos^ {2} \theta}} \\ & {\Rightarrow - 1 = x (2) - \frac {1 0 x ^ {2}}{2 v ^ {2}} (1 + 4)} \\ & {\Rightarrow - 1 = 2 x - \frac {1 0 x ^ {2}}{2 (6 0)} (5)} \\ & {\Rightarrow - 1 = 2 x - \frac {5}{1 2} x ^ {2}} \\ & {\Rightarrow - 1 2 = 2 4 x - 5 x ^ {2}} \\ & {\Rightarrow 5 x ^ {2} - 2 4 x - 1 2 = 0} \end{array}$
Q238
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle is projected towards north with speed $20 \, ms^{-1}$ at an angle $45^{\circ}$ with horizontal. Ball get horizontal acceleration of $7.5 \, ms^{-2}$ towards east due to wind. Range of ball (in meter) minus 42 m will be
Correct Answer: 8
Explanation:
Time of flight is
$T = \frac {2 u \sin \theta}{g} = 2 \sqrt {2} \mathrm{s}$
Range (along north) is
$R _ {1} = \frac {u ^ {2} \sin 2 \theta}{g} = 4 0 \mathrm{m}$
Range (along east) is
$R _ {2} = \frac {1}{2} a T ^ {2} = 3 0 \mathrm{m}$
So, range $R = \sqrt{R_{1}^{2} + R_{2}^{2}} = \sqrt{30^{2} + 40^{2}} = 50 \, m$
$T = \frac {2 u \sin \theta}{g} = 2 \sqrt {2} \mathrm{s}$
Range (along north) is
$R _ {1} = \frac {u ^ {2} \sin 2 \theta}{g} = 4 0 \mathrm{m}$
Range (along east) is
$R _ {2} = \frac {1}{2} a T ^ {2} = 3 0 \mathrm{m}$
So, range $R = \sqrt{R_{1}^{2} + R_{2}^{2}} = \sqrt{30^{2} + 40^{2}} = 50 \, m$
Q239
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
For an observer on trolley, direction of projection of particle is shown in figure, while for observer on ground ball rises vertically. Maximum height (in meter) reached by ball minus 10 m is ____
Correct Answer: 5
Explanation:
$10 - v \cos(60^{\circ}) = 0$
$\begin{array}{r l} \Rightarrow & H = \frac {v ^ {2} \sin^ {2} (6 0 ^ {\circ})}{2 g} = 1 5 \mathrm{m} \\ \Rightarrow & H - 1 0 = 5 \mathrm{m} \end{array}$
$\begin{array}{r l} \Rightarrow & H = \frac {v ^ {2} \sin^ {2} (6 0 ^ {\circ})}{2 g} = 1 5 \mathrm{m} \\ \Rightarrow & H - 1 0 = 5 \mathrm{m} \end{array}$
Q240
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
Two seconds after projection, a projectile is travelling in a direction inclined at $30^{\circ}$ with horizontal. After one more second, it is travelling horizontally. One tenth of the angle of projection (in degree) with horizontal is ____
Correct Answer: 6
Explanation:
Let angle made by $\vec{V}$ initially and after time t be $\theta$ and $\alpha$ respectively.
$\tan (3 0 ^ {\circ}) = \frac {u \sin \theta - g \times 2}{u \cos \theta}\tag{... (1}$
$\Rightarrow \tan (0 ^ {\circ}) = \frac {u \sin \theta - g \times 3}{u \cos \theta}\tag{... (2}$
$\Rightarrow \quad \theta = 6 0 ^ {\circ}$
$\Rightarrow \frac {\theta}{1 0} = 6$
$\tan (3 0 ^ {\circ}) = \frac {u \sin \theta - g \times 2}{u \cos \theta}\tag{... (1}$
$\Rightarrow \tan (0 ^ {\circ}) = \frac {u \sin \theta - g \times 3}{u \cos \theta}\tag{... (2}$
$\Rightarrow \quad \theta = 6 0 ^ {\circ}$
$\Rightarrow \frac {\theta}{1 0} = 6$
Q241
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle is projected from O on the ground with velocity $u = 5\sqrt{5} \, ms^{-1}$ at angle $\alpha = \tan^{-1}\left(\frac{1}{2}\right)$ . It strikes at a point C on a fixed plane AB having inclination of $37^{\circ}$ with horizontal as shown, then the x-coordinate of point C in meters is $(g = 10 \, \text{ms}^{-2})$
Correct Answer: 5
Explanation:
Coordinates of the point C are
$\begin{array}{r l} & C \left(\frac {1 0}{3} + x, y\right) \\ \text {Since,} \frac {y}{x} = \tan 3 7 ^ {\circ} \\ \Rightarrow & y = \frac {3}{4} x \\ \Rightarrow & u _ {y} t - \frac {1}{2} g t ^ {2} = \frac {3}{4} \left[ u _ {x} t - \frac {1 0}{3} \right] \\ \Rightarrow & t = 1. 0 6 \\ \Rightarrow & x = \frac {3}{4} \times 1 0 \times 1. 0 6 - \frac {1 0}{3} = 4. 6 4 \end{array}$
$\begin{array}{r l} & C \left(\frac {1 0}{3} + x, y\right) \\ \text {Since,} \frac {y}{x} = \tan 3 7 ^ {\circ} \\ \Rightarrow & y = \frac {3}{4} x \\ \Rightarrow & u _ {y} t - \frac {1}{2} g t ^ {2} = \frac {3}{4} \left[ u _ {x} t - \frac {1 0}{3} \right] \\ \Rightarrow & t = 1. 0 6 \\ \Rightarrow & x = \frac {3}{4} \times 1 0 \times 1. 0 6 - \frac {1 0}{3} = 4. 6 4 \end{array}$
Q242
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle is projected from ground with minimum speed required to hit a target at a height $h = 10 \, m$ at a horizontal distance $d = \sqrt{300} \, m$ as shown. Then find the time taken by particle (in seconds) to hit the target. $\left( g = 10 \, \text{ms}^{-2} \right)$
Correct Answer: 2
Explanation:
Since, $u_{\min} = \sqrt{g(h + \sqrt{d^{2} + h^{2}})} = 10\sqrt{3} \, ms^{-1}$
$\begin{array}{l l} \text {and} & \tan \theta = \frac {h + \sqrt {d ^ {2} + h ^ {2}}}{d} \\ \Rightarrow & \theta = 6 0 ^ {\circ} \\ \Rightarrow & t = \frac {d}{u \cos \theta} = \frac {1 0 \sqrt {3}}{1 0 \sqrt {3} \times \frac {1}{2}} = 2 \mathrm{s} \end{array}$
$\begin{array}{l l} \text {and} & \tan \theta = \frac {h + \sqrt {d ^ {2} + h ^ {2}}}{d} \\ \Rightarrow & \theta = 6 0 ^ {\circ} \\ \Rightarrow & t = \frac {d}{u \cos \theta} = \frac {1 0 \sqrt {3}}{1 0 \sqrt {3} \times \frac {1}{2}} = 2 \mathrm{s} \end{array}$
Q243
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A particle is projected with initial velocity $v = 10\sqrt{2} \, ms^{-1}$ as shown. After elastic collision with the inclined plane, the particle rebounds normally with the plane and retraces its path to come back at its point of projection. Then find the time in seconds in which particle returns to the point of projection. $(g = 10 \, \text{ms}^{-2})$
Correct Answer: 6
Explanation:
Since, $\tan\phi=\frac{\cot\beta}{2}$
$\Rightarrow t = \frac {2 v}{g \sqrt {1 + 3 \sin^ {2} \beta}}$
$T = \frac {2 u \sin \theta}{g} = \frac {2 \times 1 0 \times \frac {1}{2}}{1 0} = 1 \mathrm{s}$
$\Rightarrow t = \frac {2 v}{g \sqrt {1 + 3 \sin^ {2} \beta}}$
$T = \frac {2 u \sin \theta}{g} = \frac {2 \times 1 0 \times \frac {1}{2}}{1 0} = 1 \mathrm{s}$
Q244
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6. INTEGER/NUMERICAL ANSWER TYPE QUESTIONS
Numerical
A football is thrown with a velocity of $10 \, ms^{-1}$ at an angle of $30^{\circ}$ above the horizontal. What will the time of flight, in seconds? ( $g = 10 \, ms^{-2}$ )
Correct Answer: 1
Explanation:
1