Question

When a golf club hits a 0.0459 kg ball at rest, it exerts a 2380 N force for 0.00100 s. What is the speed of the ball afterwards? (Unit = m/s)

Answer 1

2.38

Step-by-step explanation:

F = dp/dt --> dp = F*dt ---> p = F*t if F is constant over time so

p = 2380N* 0.001 s = 2.38 kg-m/s

Answer 2

51.85m/s

Step-by-step explanation:

Given parameters:

Mass of ball = 0.0459kg

Force = 2380N

Time taken = 0.001s

Unknown:

Speed of the ball afterwards = ?

Solution:

To solve this problem, we use Newton's second law of motion:

F = m x
`(v - u)/(t)`

F is the force

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

2380 = 0.0459 x
`(v- 0)/(0.001)`

0.0459v = 2.38

v = 51.85m/s