Physics problem newton's third law?

Question

A 60.0 g ball of clay is thrown horizontally with a speed of 9.00 m/s toward a 910 g block resting on a frictionless surface. It hits the block and sticks. The clay exerts a constant force on the block during the 10.0 ms it takes the clay to come to rest relative to the block. After 10.0 ms, the block and the clay are sliding along the surface as a single system.
What is their speed after the collision?
What is the force of the clay on the block during the collision?
What is the force of the block on the clay?

Answer 1

Momentum is always conserved, so mv = MV. Initial momentum is 9*60 = 540 m-kg/s. Then
970*V = 540 or V = .5567 m/s

The accelleration of the block is .5567/.01 = 55.67 m/s² and F= ma so F = 910*55.67 = 50659.7 N

The block and the clay are a coupled 'force pair' so the force on the clay is the same 50659.7 N.

In fact, if you work out the final velocity of the clay after a -decelleration- of a = F/m (using 50659.7 N and 60 kg) for a period of .01s (it's v = v0 - at) you'll find that the clay is travelling at the same speed (.5567 m/s) as the block (which is *exactly* what you'd expect ☺)


Doug