Quick Question - Conservation of Momentum

[Dooga Blackrazor]

ball with a mass of 0.10 kg has a velocity of 7 m/s. It strikes a concrete wall perpendicularly and bounces straight back with velocity of 4 m/s. The change in the ball's momentum was equal to

Response Answer Choices
a. 0.30 kg m/s
b. 0.10 kg m/s
c. 0.40 kg m/s
d. 1.1 kg m/s

0.1 x 7 and .4 x .1 = .7 and .4 and subtracting them gives .30 kg m/s, correct?

 

[Dooga Blackrazor]

Nevermind, I found the answer (1.1). Silly mistake.

 

[quicknote]

Your calculation is correct. The change in momentum is equal to the initial momentum minus the final momentum, which in this case is 0.7 kg m/s - 0.4 kg m/s = 0.3 kg m/s. This value represents the conservation of momentum, which states that the total momentum of a system remains constant unless acted upon by an external force. In this scenario, the external force is the impact with the concrete wall, causing the ball's momentum to change.