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

Answer:

51.85m/s

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 $\frac{v - u}{t}$  

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 $\frac{v- 0}{0.001}$  

        0.0459v  = 2.38

                   v = 51.85m/s

Answer 2

Answer:

2.38

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