Laws of Motion

Apply conservation of linear momentum. Total momentum before explosion = total momentum after explosion

$0 = {m \over 5}{v_1}\widehat i + {m \over 5}{v_2}\widehat j + {{3m} \over 5}{\overrightarrow v _3}$

${{3m} \over 5}{\overrightarrow v _3} = - {m \over 5}\left[ {{v_1}\widehat i + {v_2}\widehat j} \right]$

${\overrightarrow v _3} = {{ - {v_1}} \over 3}\widehat i - {{{v_2}} \over 3}\widehat j$
     $ \therefore $ ${v_1} = {v_2} = 30$ m/sec

${\overrightarrow v _3} = - 10\widehat i - 10\widehat j;{v_3} = 10\sqrt 2 $ m/sec

Load W = Mg = 75 × 10 = 750 N

Effort (P) = 250 N

$ \therefore $ Mechanical advantage
$ = {{load} \over {effort}} = {W \over P} = {{750} \over {250}} = 3$

Velocity ratio
$ = {{distance\,travelled\,by\,effort} \over {distance \,travelled\,by\,load}} = 4$

Efficiency $(\eta) = {{Mechanical\,advantage} \over {Velocity\,ratio}}$

$ = \left( {3/4} \right) \times 100 = 75\% $

Let T be the tension in the string.

$ \therefore $ 10g – T = 10a     ....(i)

T – 5g = 5a             ....(ii)

Adding (i) and (ii),

$5g = 15a \Rightarrow a = {g \over 3}m/{s^2}$

Change in momentum = 2 × 3 × 10 × sin60°

= $60 \times {{\sqrt 3 } \over 2}$

Force = Change in momentum/Impact time

$ = {{30\sqrt 3 } \over {0.2}} = 150\sqrt 3 N$