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2 years ago

b. What would the momentum be if the mass of the bowling ball were doubled and its velocity still was 3 m/s?

Physics

1 answer:

umka2103 [35]2 years ago

6 0

Answer:

Twice.

Explanation:

The momentum of an object is given by :

p = mv

Where

m is mass and v is the velocity

If the mass of the ball were doubled, m'=2m and v'=v=3 m/s

New momentum,

p'=m'v'

p'=2m × v

p'=2mv

p'=2p

So, the new momentum becomes twice the initial momentum.

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The first car has twice the mass of a second car, but only half as much kinetic energy. When both cars increase their speed by 9

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Speed of the car 2 = $V_2=17.96m/s$

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also,

$\frac{1}{2}M_1(V_1+9.0)^2=\frac{1}{2}M_2(V_2+9.0)^2$

$\frac{1}{2}2M_2(V_1+9.0)^2=\frac{1}{2}M_2(2V_1+9.0)^2$

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$(\sqrt{2}V_1+ \sqrt{2}\times 9.0)=(2V_1+9.0)$

$(\sqrt{2}V_1+ 12.72)=(2V_1+9.0)$

$(2V_1-\sqrt{2}V_1)=(12.72-9.0)$

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Hence,

Speed of car 1 = $V_1=8.98m/s$

Speed of car 2 = $V_2=17.96m/s$

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b. What would the momentum be if the mass of the bowling ball were doubled and its velocity still was 3 m/s?​

b. What would the momentum be if the mass of the bowling ball were doubled and its velocity still was 3 m/s?