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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

Best Answer 

 #1
avatar+2786 
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\(\text{conservation of momentum is a vector equation}\\ \text{assume the velocity of the bowling ball is }v_i = -2\hat{i} \\ \text{moving left along the x-axis if you like}\\ \text{when it recoils it's given to have the same speed}\\ \text{but now it's velocity will be }v_f = +2\hat{i}\\ \text{it's momentum of recoil is just }\\ m\vec{v} = 8kg \cdot (2\hat{i}~m/s) =16 \hat{i}~kg\cdot m/s\\ \text{the change is just twice this since it was originally exactly negative this}\\ \text{so }\Delta mv = 32\hat{i}~kg\cdot m/s\)

 

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

Rom  Nov 5, 2018
edited by Rom  Nov 5, 2018
 #1
avatar+2786 
+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
edited by Rom  Nov 5, 2018

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