A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?
Answer
= 9.25 m/s
Explanation
The Newton's second law of motion states that, the change in momentum is directly propotional to the force producing it and it takes place in the direction of force.
F = ma
f = m(v-u)/t
ft = m(v-u)
∴ 55 × 45/1000 = 0.060(v - -32)
2.475 = 0.06(v + 32)
2.475/0.6 = v + 32
41.25 = v + 32
v = 41.25 -32
= 9.25 m/s