Question

A 0.5 kg ball is dropped vertically onto a floor, hitting with a speed of 20 m/s. It rebounds with a speed of 15 m/s immediately after the collision. (a) What is the impulse in N-s acting on the ball during the contact

Answer 1

Answer:

2.499Ns

Explanation:

Impulse is defined as change I'm momentum of a body. It is expressed as;

According to Newton's second law;

F = ma

F = m(v-u)/t

Ft = m(v-u) = Impulse

Impulse = Ft

Impulse = m(v-u)

F is the force acting on the ball

t is the time taken

m is the mass of the body

v is the final velocity

u is the initial velocity

To get impulse, we need to get time t first using the equation of motion v = u+gt

20 = 15+(9.8)t

20-15 = 9.8t

5 = 9.8t

t = 5/9.8

t = 0.51secs

Since Impulse = Ft

F = mg = 0.5(9.8)

F = 4.9N

Impulse = 4.9×0.51

Impulse acting on the ball during contact = 2.499Ns