Question

A moving body of mass m and velocity 3km/hr collides with an arrest body of mass 2m and sticks to it. Now the combined mass starts to move, what will be the combined velocity?
$(a){\text{ 4Km/hr}} \\\ (b){\text{ 1Km/hr}} \\\ (c){\text{ 2Km/hr}} \\\ (d){\text{ 3Km/hr}} \\\$

Answer 1

- Hint: In this question use the mass ${m_1} = m$ Kg and ${u_1} = 3$ km/h for body A which is moving and ${m_2} = 2m$ kg and ${u_2} = 0$ m/s for body B as this is at rest. Application of conservation of momentum before and after the collision of the system of the two masses will help finding out the combined velocity of the system.

Complete step-by-step solution -

Given data:
Before colliding, let us assume that body (A) moves in the positive direction of x-axis and body B is at rest.
Body A: ${m_1} = m$ Kg and ${u_1} = 3$ km/h
Body B: ${m_2} = 2m$ Kg and ${u_2} = 0$ m/s
After colliding both bodies stick together.
Let the velocity of the combined body be v km/h.
Combined body A and B: ${m_1} = m + 2m = 3m$ Kg and $v =$? Km/h (continue moving in same direction)
Now according to conservation of momentum, momentum before colliding is equal to the momentum after colliding.
And we all know momentum is the product of mass and velocity.
Therefore, sum of momentum of body A and body B before collision = momentum of combined body after collision.
$\Rightarrow {m_1}{u_1} + {m_2}{u_2} = \left( {{m_1} + {m_2}} \right)v$
Now substitute the values we have,
$\Rightarrow \left( m \right)\left( 3 \right) + \left( {2m} \right)\left( 0 \right) = \left( {3m} \right)\left( v \right)$
Now simplify this we have,
$\Rightarrow 3m = 3mv$
Now cancel out the common terms we have,
$\Rightarrow v = 1$ Km/h.
So the velocity of the combined body after colliding comes positive so the combined body moves in the positive direction of x-axis or they move in the same direction of motion before collision.
And the velocity of the combined body is 1Km/h.
Hence option (B) is the correct answer.

Note – Since the bodies stick together after the collision therefore it implies that the collision was perfectly inelastic that is coefficient of restitution in this case will be 0. The coefficient of restitution is always in between 0 to 1, that is $0 < e < 1$.