JEE MAIN - Physics (2008 - No. 29)
A block of mass $$0.50$$ $$kg$$ is moving with a speed of $$2.00$$ $$m{s^{ - 1}}$$ on a smooth surface. It strike another mass of $$1.0$$ $$kg$$ and then they move together as a single body. The energy loss during the collision is :
$$0.16J$$
$$1.00J$$
$$0.67J$$
$$0.34$$ $$J$$
Explanation
Let $$m$$ = 0.50 kg and $$M$$ = 1.0 kg
Initial kinetic energy of the system when 1 kg mass is at rest,
$$K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 \right)^2}$$
$$ = {1 \over 2} \times 0.5 \times 2 \times 2 + 0 = 1J$$
For collision, applying conservation of linear momentum
$$\,\,\,\,\,\,\,\,\,\,\,\,m \times u = \left( {m + M} \right) \times v$$
$$\therefore$$ $$0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = {2 \over 3}m/s$$
Final kinetic energy of the system is
$$K.{E_f} = {1 \over 2}\left( {m + M} \right){v^2}$$
$$ = {1 \over 2}\left( {0.5 + 1} \right) \times {2 \over 3} \times {2 \over 3} = {1 \over 3}J$$
$$\therefore$$ Energy loss during collision
$$ = \left( {1 - {1 \over 3}} \right)J = 0.67J$$
Initial kinetic energy of the system when 1 kg mass is at rest,
$$K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 \right)^2}$$
$$ = {1 \over 2} \times 0.5 \times 2 \times 2 + 0 = 1J$$
For collision, applying conservation of linear momentum
$$\,\,\,\,\,\,\,\,\,\,\,\,m \times u = \left( {m + M} \right) \times v$$
$$\therefore$$ $$0.5 \times 2 = \left( {0.5 + 1} \right) \times v \Rightarrow v = {2 \over 3}m/s$$
Final kinetic energy of the system is
$$K.{E_f} = {1 \over 2}\left( {m + M} \right){v^2}$$
$$ = {1 \over 2}\left( {0.5 + 1} \right) \times {2 \over 3} \times {2 \over 3} = {1 \over 3}J$$
$$\therefore$$ Energy loss during collision
$$ = \left( {1 - {1 \over 3}} \right)J = 0.67J$$
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