A 100 g ball of clay is thrown horizontally with a speed of 50.0 m/s toward a 900 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 the force of the clay on the block during the collision?
A)450 N
B)4500 N
C)500 N
D)2250 N

Question

Quizplus · Accepted Answer

The impulse on the block is equal to the change in momentum of the block-clay system:$$J = mv_f - mv_i$$$$J = (1000 g)(v_f - 0) = 1 kg(v_f)$$The force of the clay on the block is equal to the rate of change of momentum of the block-clay system:$$F = \frac{J}{t} = \frac{1 kg(v_f)}{0.01 s} = 1000 v_f$$The speed of the block-clay system after the collision is:$$v_f = \frac{mv_i}{m_1 + m_2} = \frac{(100 g)(50 m/s)}{1000 g} = 5 m/s$$Therefore, the force of the clay on the block during the collision is:$$F = 1000 v_f = 1000 (5 m/s) = 5000 N$$

A 100 G Ball of Clay Is Thrown Horizontally with a Speed