Question

A bullet weighing $10g$ and moving at $300m/s$ strikes a $5kg$ ice and drops dead. The ice block is sitting on a frictionless level surface. The speed of the block after the collision is
A. ${\text{6cm/s}}$
B. ${\text{6m/s}}$
C. ${\text{60cm/s}}$
D. ${\text{60m/s}}$

Answer 1

Hint: First convert all the units of the given terms in the same system of units. Here use the basic law for the conservation of momentum and the velocity. Momentum in an isolated system remains constant. The law of the momentum states that momentum can never be created nor destroyed. It states that momentum of an object before collision and after collision remains the constant.

Complete step by step answer:
Let us suppose that mass be ${m_1} = 10g = 10 \times {10^{ - 3}}kg$
Velocity of the bullet be ${v_1} = 300m/s$
Velocity of the ice is ${v_2} = 0m/s$
Mass of the ice, is ${m_2} = 5kg$
In any isolated system, the total momentum of any object before the collision is equal to the total momentum after the collision.
Given that - A bullet weighing $10g$ and moving at $300m/s$ strikes a $5kg$ ice and drops dead
 ${m_1}{v_1} + {m_2}{v_2} = {m_2}v$
Place the known values in the above equation –
$10 \times {10^{ - 3}}(300) + 5(0) = 5v$
Simplify the above equation and make unknown “v” the subject –
$3000 \times {10^{ - 3}} = 5v \\$
$\implies 3 = 5v \\$
$\implies v = \dfrac{3}{5} \\$
$\therefore v = 0.6m/s \\ $
Therefore, the speed of the block after the collision is $0.6m/s = 0.6 \times 100cm/s = 60cm/s$

So, the correct answer is “Option C”.

Note:
The law of momentum is applicable to all the interactions, collisions caused by the explosive forces and applicable to all the systems no matter how complicated the force is between the particles. It can also be used and applicable where Newton’s laws do not hold. Remember all the basic laws to solve these types of equations.