Question

A bullet of mass 10 g moving with 300 m/s hits a block of ice of mass 5 kg and drops dead. The velocity of the ice is
A. 50cm/s
B. 60cm/s
C. 40 cm/s
D. 200 cm/s

Answer 1

Hint: It is based on the principle of law of conservation of momentum. This law is applicable for an isolated system which states that the momentum can neither be created nor be destroyed.

Formula Used:
The formula for maw of conservation of momentum will be used. Momentum is the product of mass and velocity of the particle in the system. Since it has both magnitude and direction, it is a vector quantity. Mathematically law of conservation of momentum is written as,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]

Complete step by step solution:
Given that the mass of the bullet is \[{m_1} = 10g = {10^{ - 3}}kg\]
Also, velocity of the bullet is given as \[{v_1} = 300m/s\]
Mass of the ice block is given \[{m_2} = 5kg\]
Velocity of the ice block \[{v_2} = ?\]
Since in an isolated system, the momentum of the system before collision and after collision remains constant, so it can be written that,
\[{m_1}{u_1} + {m_2}{u_2} = {m_1}{v_1} + {m_2}{v_2}\]

Substituting all the values in the above equation and solving for the value of velocity we get,
\[({10^{ - 2}} \times 300) + (5 \times 0) = ({10^{ - 2}} \times 0) + 5{v_2}\]
\[\Rightarrow 3 = 5{v_2}\]
\[\Rightarrow {v_2} = \dfrac{3}{5}\]
\[\Rightarrow {v_2} = 0.6\,m/s\]
Converting it into cm, we get
\[{v_2} = 60\,cm/s\]
Therefore, the velocity of the ice block when the other block of mass strikes with it is 60 cm/s.

Hence, option B is the correct answer.

Note: It is to be noted that according to the law of conservation of momentum, the momentum of an object before the collision and after the collision, remains the same. It can be applied if the objects are colliding with each other, exploding or interacting with each other. In case, if Newton’s law can not be applied to the system, the law of conservation of momentum is also applicable in that case.