Question

Physics! Someone help me! Its due today please!

Assume a 9.0 kg bowling ball moving at 5 m/s bounces off a spring at 2 m/s.


(a) What is its change in velocity of the bowling ball?

(b) What is its change of momentum of the ball?

(c) What is the impulse exerted on of the ball?

(d) If the interaction with the spring occurs in 0.3 s, calculate the average force the spring exerts on the ball.

Answer 1

Step-by-step explanation:

We have,

Mass of ball is 9 kg

Initial speed, u = 5 m/s

Final speed, v = -2 m/s (negative as it bounces off)

(a) The change in velocity of the bowling ball is :

$\Delta v=v-u\\\\\Delta v=-2-5\\\\\Delta v=-7\ m/s$

(b) Change of momentum of the ball is :

$p=m\Delta v\\\\p=90\times (-7)\\\\p=-630\ kg-m/s$

|p| = 630 kg-m/s

(c) Impulse momentum theorem states that the change in momentum of the ball is equal to the impulse exerted on the ball. So, impulse is 630 kg-m/s.

(d) Impulse is also given by :

$J=F\times t\\\\F=\dfrac{J}{t}\\\\F=\dfrac{630}{0.3}\\\\F=2100\ N$