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riadik2000 [5.3K]
3 years ago
12

A 20 g bullet is shot from a 10 kg gun with a velocity of 400 m/s. What

Physics
1 answer:
Nady [450]3 years ago
3 0

Answer: -.80m/s

Explanation:

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Answer:

  Second  projectile is 1.4 times faster than first projectile.

Explanation:

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U= \dfrac{(m + M)\times V}{m}

Now Since this projectile + pendulum system rises to height 'h', So using energy conservation:

KEi + PEi = KEf + PEf

PEi = 0, at reference point

KEf = 0, Speed of system zero at height 'h'

KEi = \dfrac{(m + M)\times V^2}{2}

PEf = (m + M) g h

So,

\dfrac{(m + M)\times V^2}{2} + 0 = 0+ (m + M) g h

V =\sqrt {2gh}

So from above value of V

Initial velocity of projectile =U

U=\dfrac{(M+m)\sqrt{2gh}}{m}

Now Since mass of projectile and pendulum are constant, So Initial velocity of projectile is proportional to the square root of height swung by pendulum.

Which means

\dfrac{U_2}{U_1}=\sqrt{\dfrac{h_2}{h_1}}

U_2=\sqrt{\dfrac{h_2}{h_1}}\times U_1

U_2=\sqrt{\dfrac{5.2}{2.6}}\times U_1

U₂ = 1.41 U₁

Therefore we can say that ,Second  projectile is 1.4 times faster than first projectile.

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0.4m/s in the opposite direction of the ball.

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