Assume an 8 kg bowling ball moving at 2 m/s bounces off a spring at the same speed that it had before bouncing. a) what is its momentum of recoil? b) what is its change in momentum? (Hint: What is the change in temperature when something goes from 1 to -1 degrees? c) If the interaction with the spring occurs in 0.5s, calculate the average force the spring exerts on it.

RainbowPanda
Nov 5, 2018

#1**+2 **

\(\text{conservation of momentum is a vector equation}\\ \text{assume the velocity of the bowling ball is }v_i = -2\hat{i} \\ \text{moving left along the x-axis if you like}\\ \text{when it recoils it's given to have the same speed}\\ \text{but now it's velocity will be }v_f = +2\hat{i}\\ \text{it's momentum of recoil is just }\\ m\vec{v} = 8kg \cdot (2\hat{i}~m/s) =16 \hat{i}~kg\cdot m/s\\ \text{the change is just twice this since it was originally exactly negative this}\\ \text{so }\Delta mv = 32\hat{i}~kg\cdot m/s\)

\(F = \dfrac{d~mv}{dt} \\ \text{and in this case this means}\\ F = \dfrac{\Delta mv}{t} \\ F = \dfrac{32\hat{i}}{0.5} = 64\hat{i} ~N\\ \text{note that }F \text{ is a vector. It has direction as expected}\)

Rom
Nov 5, 2018

#1**+2 **

Best Answer

\(\text{conservation of momentum is a vector equation}\\ \text{assume the velocity of the bowling ball is }v_i = -2\hat{i} \\ \text{moving left along the x-axis if you like}\\ \text{when it recoils it's given to have the same speed}\\ \text{but now it's velocity will be }v_f = +2\hat{i}\\ \text{it's momentum of recoil is just }\\ m\vec{v} = 8kg \cdot (2\hat{i}~m/s) =16 \hat{i}~kg\cdot m/s\\ \text{the change is just twice this since it was originally exactly negative this}\\ \text{so }\Delta mv = 32\hat{i}~kg\cdot m/s\)

\(F = \dfrac{d~mv}{dt} \\ \text{and in this case this means}\\ F = \dfrac{\Delta mv}{t} \\ F = \dfrac{32\hat{i}}{0.5} = 64\hat{i} ~N\\ \text{note that }F \text{ is a vector. It has direction as expected}\)

Rom
Nov 5, 2018