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A machine gun has a mass of 20 kg fires 35 g bullets at the rate of 400 bullets per second with a speed of 400ms-1 .What force must be applied to the gun to keep in the position?
- a)5800 N
- b)5600 N
- c)560 N
- d)56 N
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
A machine gun has a mass of 20 kg fires 35 g bullets at the rate of 40...
To conserve the momentum each time a single bullet is fired,
the reverse speed gained by the gun from one bullet is
V = 400 X .035 / 20
= 0.7 m/s
Thus total speed gained in a second is = 0.7 X 400 = 280 m/s
As total speed is gained in one second only the acceleration produced = 280 m/s2
Thus total force applied on the gun by the bullets = 20 x 280
= 5600 N
the reverse speed gained by the gun from one bullet is
V = 400 X .035 / 20
= 0.7 m/s
Thus total speed gained in a second is = 0.7 X 400 = 280 m/s
As total speed is gained in one second only the acceleration produced = 280 m/s2
Thus total force applied on the gun by the bullets = 20 x 280
= 5600 N
Most Upvoted Answer
A machine gun has a mass of 20 kg fires 35 g bullets at the rate of 40...
Given:
Mass of the machine gun, m = 20 kg
Mass of each bullet, m_b = 35 g = 0.035 kg
Rate of firing, n = 400 bullets/second
Speed of each bullet, v = 400 m/s
To find: Force required to keep the gun in position
Formula used:
Force = rate of change of momentum
Calculation:
1. Mass of each bullet, m_b = 0.035 kg
2. Total mass of n bullets fired in 1 second =
Mass of one bullet x number of bullets fired in 1 second
= m_b x n
= 0.035 x 400
= 14 kg
3. Initial momentum of the bullets, p_i =
Total mass of n bullets fired in 1 second x velocity of each bullet
= 14 x 400
= 5600 kg m/s
4. Final momentum of the bullets, p_f = 0 (since the bullets stop after firing)
5. Change in momentum, Δp = p_f - p_i = - p_i = -5600 kg m/s
6. Rate of change of momentum, Δp/Δt = -5600 kg m/s
7. Force required to keep the gun in position, F =
Rate of change of momentum x mass of the gun
= -5600 x 20
= -112000 N
(Note that the negative sign indicates that the force is acting in the opposite direction to the motion of the bullets)
Therefore, the force required to keep the gun in position is 5600 N (option B).
Mass of the machine gun, m = 20 kg
Mass of each bullet, m_b = 35 g = 0.035 kg
Rate of firing, n = 400 bullets/second
Speed of each bullet, v = 400 m/s
To find: Force required to keep the gun in position
Formula used:
Force = rate of change of momentum
Calculation:
1. Mass of each bullet, m_b = 0.035 kg
2. Total mass of n bullets fired in 1 second =
Mass of one bullet x number of bullets fired in 1 second
= m_b x n
= 0.035 x 400
= 14 kg
3. Initial momentum of the bullets, p_i =
Total mass of n bullets fired in 1 second x velocity of each bullet
= 14 x 400
= 5600 kg m/s
4. Final momentum of the bullets, p_f = 0 (since the bullets stop after firing)
5. Change in momentum, Δp = p_f - p_i = - p_i = -5600 kg m/s
6. Rate of change of momentum, Δp/Δt = -5600 kg m/s
7. Force required to keep the gun in position, F =
Rate of change of momentum x mass of the gun
= -5600 x 20
= -112000 N
(Note that the negative sign indicates that the force is acting in the opposite direction to the motion of the bullets)
Therefore, the force required to keep the gun in position is 5600 N (option B).
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Community Answer
A machine gun has a mass of 20 kg fires 35 g bullets at the rate of 40...
We can also apply this formula F=mnv m=mass of bullent in kg n=no. of bullet per second v=velocity of bullet in m/s F=0.035×400×400 F=5600N
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A machine gun has a mass of 20 kg fires 35 g bullets at the rate of 400 bullets per second with a speed of 400ms-1 .What force must be applied to the gun to keep in the position?a)5800 Nb)5600 Nc)560 Nd)56 NCorrect answer is option 'B'. Can you explain this answer?
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