Work, Energy and Power

It is observed that if mass colloids with spring in such case, spring will be compressed, so kinetic energy of mass will be equal to that of springs potential energy.

Further, if y is the compression, then

${1 \over 2}m{v^2} = {1 \over 2}k{y^2}$

$ \Rightarrow {1 \over 2}\left( {0.5} \right)\left( {1.5} \right) = {1 \over 2}\left( {50} \right){y^2}$

$ \therefore $ y = 0.15 m

Kinetic energy = ${{{p^2}} \over {2m}}$

${{{E_1}} \over {{E_2}}} = {{p_1^2/2{m_1}} \over {p_2^2/2{m_2}}} \Rightarrow {{{E_1}} \over {{E_2}}} = {{{m_2}} \over {{m_1}}}$ as m₁ > m₂

$ \therefore $ E₁ < E₂

From the question, as height is same for both balls, so their velocities on reaching the ground will be same.

Hence in such case kinetic energy ∝ mass

So, ${{K.{E_1}} \over {K.{E_2}}} = {{{m_1}} \over {{m_2}}} = {1 \over 2}$

W_mg = $\Delta $K

$ \Rightarrow $ -mg$\ell $ = ½ mv² – ½ mu²
or, mv² = m(u² – 2g$\ell $]
or, $v = \sqrt {{u^2} - 2g\ell } \widehat j$

$\overrightarrow u = u\widehat i$

$ \therefore \overrightarrow v - \overrightarrow u = \sqrt {{u^2} - 2g\ell } \widehat j - u\widehat i$

$ \therefore \left| {\overrightarrow v - \overrightarrow u } \right| = {\left[ {\left( {{u^2} - 2g\ell } \right) + {u^2}} \right]^{{1 \over 2}}}$

$ = \sqrt {2\left( {{u^2} - g\ell } \right)} $

If k be the spring constant, then

$U = {1 \over 2} \times k \times {\left( 2 \right)^2} = 2k$

${U_{final}} = {1 \over 2} \times k \times {\left( {10} \right)^2} = 50k$

$ \Rightarrow {U \over {{U_{final}}}} = {{2k} \over {50k}} = {1 \over {25}}$

$ \Rightarrow {U_{final}} = 25U$

Kinetic energy of the ball = K and angle of projection ($\theta $) = 45^o

Velocity of the ball at the highest point = v cos $\theta $

$ = v\,\,\cos \,\,{45^o} = {v \over {\sqrt 2 }}$

Therefore kinetic energy of the ball

$ \Rightarrow $ ${1 \over 2}m \times \left( {{v \over {\sqrt 2 }}} \right) = {1 \over 4}m{v^2} = {K \over 2}$

Energy = ${1 \over 2}K{x^2} = {1 \over 2}{{{F^2}} \over K}$

${{{K_A}} \over {{K_B}}} = 2$

$ \therefore $ ${{{E_A}} \over {{E_B}}} = {1 \over 2} \Rightarrow {E_B} = 2{E_A}$

Drop in P.E. = maximum K.E.

mg(2 – 0.75) = 1mv²

$ \Rightarrow $ v = $\sqrt {2g\left( {1.25} \right)} = 5\,m/s$

P = $\overrightarrow F .\overrightarrow v = \left( {60\widehat i + 15\widehat j - 3\widehat k} \right).\left( {2\widehat i - 4\widehat j + 5\widehat k} \right)$

= 120 – 60 – 15 = 45 watts.

Velocity after 5 sec,

v = u – gt = 100 – 10 × 5 = 50 m/s

By conservation of momentum

1 × 50 = 0.4 × (–25) + 0.6 × v'

60 = 0.6 × v' $ \Rightarrow $ v' = 100 m/s upwards