A clay blob of mass m1, initially at rest, is pushed across a frictionless surface with constant force F for a distance d. It then hits and sticks to a second clay blob of mass m2 that is at rest.

Find an expression for their speed after the collision.

Express your answer in terms of the variables m1, m2, d, and F.

Answer: V-final = {F*{d/v¹}} / {m1 + m2}

Explanation: From Newton's second law we know that impulse that is force multiplied by time is equal to change in momentum.

Therefore the momentum before collision would be

F*t

But time t = distance (d) / velocity (V1)

t = d/V1

Momentum before collision

=F * {d/v1}.

Also, from the question we were told after collision the two balls stuck together. That statement means that they move with a common (same) velocity after collision.

Therefore,

Momentum after collision

= {m1+m2}*V-final

From the law of conservation of momentum which state that the sum of momentum before collision is EQUAL to the momentum after collision. We have that,

F*{d/v1} = {m1+m2}*V-final

Making V - final subject of the formula we have that,

V-final = {F*{d/v1}} / {m1+m2}

NOTE: The momentum of a body is the product of it's mass and velocity. That is M * V.