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GenaCL600 [577]

2 years ago

5

A person holding a 15.0 kg containing one 50.0 g bullet is riding on a train that is traveling at 75.0 km/h east. If the man fir

es the gun and the bullet moves with a velocity of 350 m/s east(relative to the train), what is the velocity of the gun relative to the ground

1 answer:

Lana71 [14]2 years ago

8 0

Answer:

The velocity of the gun relative to the ground is 19.66 m/s

Explanation:

Given data,

The mass of the gun, M = 15.0 kg

The mass of the bullet, m = 50 g

The velocity of the train, v = 75 km/h

= 20.83 m/s

The velocity of bullet relative to train, V' = 350 m/s

The velocity of bullet relative to ground, V = 350 + 20

= 370 m/s

According to the law of conservation of momentum,

Mv' + mV' = 0

$v' = -\frac{mV'}{M}$

$v' = -\frac{0.050\times 350}{15}$

= -1.17 m/s

Therefore, the velocity of the gun with,

v₀ = V + v'

= 20.83 - 1.17

= 19.66 m/s

Hence, the velocity of the gun relative to the ground is 19.66 m/s

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Given that,

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2 years ago

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Answer:

Part a)

$x = 15.76 m$

Part b)

$y = 7.94 m$

Part c)

$x = 26.16 m$

Part d)

$y = 7.49 m$

Part e)

$x = 83.23 m$

Part f)

$y = -75.6 m$

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

$v_x = 19.9 cos39.9$

$v_x = 15.3 m/s$

similarly we have

$v_y = 19.9 sin39.9$

$v_y = 12.76 m/s$

Part a)

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$x = v_x t$

$x = (15.3)(1.03)$

$x = 15.76 m$

Part b)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.03) - 4.9(1.03)^2$

$y = 7.94 m$

Part c)

Horizontal displacement in 1.71 s

$x = v_x t$

$x = (15.3)(1.71)$

$x = 26.16 m$

Part d)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(1.71) - 4.9(1.71)^2$

$y = 7.49 m$

Part e)

Horizontal displacement in 5.44 s

$x = v_x t$

$x = (15.3)(5.44)$

$x = 83.23 m$

Part f)

Vertical direction we have

$y = v_y t - \frac{1]{2}gt^2$

$y = (12.76)(5.44) - 4.9(5.44)^2$

$y = -75.6 m$

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2 years ago