Question 221: Centre of Mass and Rotational Motion

Question

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 :

Accepted Answer

Let $$m$$ = 0.50 kg and $$M$$ = 1.0 kg
Initial kinetic energy of the system when 1 kg mass is at rest,
$$K.{E_i} = {1 \over 2}m{u^2} + {1 \over 2}M{\left( 0 ight)^2}$$
$$ = {1 \over 2} 	imes 0.5 	imes 2 	imes 2 + 0 = 1J$$
For collision, applying conservation of linear momentum
$$\,\,\,\,\,\,\,\,\,\,\,\,m 	imes u = \left( {m + M} ight) 	imes v$$
$$	herefore$$ $$0.5 	imes 2 = \left( {0.5 + 1} ight) 	imes v \Rightarrow v = {2 \over 3}m/s$$
Final kinetic energy of the  system is
$$K.{E_f} = {1 \over 2}\left( {m + M} ight){v^2}$$
$$ = {1 \over 2}\left( {0.5 + 1} ight) 	imes {2 \over 3} 	imes {2 \over 3} = {1 \over 3}J$$
$$	herefore$$ Energy loss during collision
$$ = \left( {1 - {1 \over 3}} ight)J = 0.67J$$

Answer

$$0.16J$$

Answer

$$1.00J$$

Answer

$$0.67J$$

Answer

$$0.34$$  $$J$$

Centre of Mass and Rotational Motion

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 :

Choose the correct answer: