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liq [111]
3 years ago
11

A bowling ball has an initial momentum of +30 kg m/s and hits a stationary bowling pin. After the collision, the bowling ball le

aves with a momentum of +13 kg • m/s
Physics
1 answer:
mr Goodwill [35]3 years ago
4 0

Answer:

+17 kg m/s

Explanation:

Question is missing. Found it on google:

"<em>What is the magnitude of the final momentum of the bowling pin if it has a mass of 1.5 kg</em>?"

Solution:

we can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system must be conserved, so we can write:

p_i = p_f\\p_b+p_p=p_b'+p_p'

where

p_b = +30 kg m/s is the momentum of the ball before the collision

p_p = 0 is the momentum of the pin before the collision (zero because the pin is stationary)

p_b' = +13 kg m/s is the momentum of the ball after the collision

p_p' is the momentum of the pin after the collision

Solving the equation for p_p', we find:

p_p' = p_b + p_p - p_b' = +30 +0 -(+13)=+17 kg m/s

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\dfrac{dz}{dt}=0.65\ ft/s

Explanation:

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