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3 years ago

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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

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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Answer:

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Explanation:

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Explanation:

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So, $I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}$

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in a closed system three objects have the following momentum: 11 kg* m/s, -65 kg*m/s and -100 kg m/s. the objects collide and mo

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Explanation:

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It would mean that, after collision, momentum of the system is equal to the initial momentum.

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Explanation:

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3 years ago