A 25 g bullet with a muzzle velocity of 300 m/s is fired into a board 4.0 cm thick. A). In a hard board, the bullet penetrates 3.0 cm and comes to a stop. What is the average force exerted on the bullet by the board?
B). For a soft board, the bullet goes through the board and emerges with a speed of 50 m/s. What was the average force exerted on the bullet by the board.
Answers:
A= 37,500 N
B= 27344 N
Show work
momentum = m v = .025 * 300
stops in .03 meters
average speed during stop = 150 m/s
time to stop = .03 meters/150 m/s
force = change in momentum/time
= .025 * 300 / (.03/150)
= 37,500 N
do B the same way
Maybe we could just get rid of guns... ¯\_(ツ)_/¯
A) To find the average force exerted on the bullet by the board in scenario A where the bullet penetrates 3.0 cm and comes to a stop, we can use the principle of work and energy.
The work done on an object is equal to the product of the force applied and the distance it moves in the direction of the force. In this case, the force exerted by the board decelerates the bullet until it comes to a stop. The work done by the board is equal to the change in kinetic energy of the bullet.
The initial kinetic energy of the bullet is given by:
KE_initial = (1/2) * m * v_initial^2
where m is the mass of the bullet (25 g = 0.025 kg) and v_initial is the initial velocity of the bullet (300 m/s).
The final kinetic energy of the bullet when it comes to a stop is zero.
Therefore, the work done by the board is equal to the initial kinetic energy of the bullet:
Work = KE_final - KE_initial
= 0 - (1/2) * m * v_initial^2
= - (1/2) * 0.025 kg * (300 m/s)^2
Now, the work done by the board is equal to the force exerted by the board (F) multiplied by the distance the bullet penetrates into the board (d = 3.0 cm = 0.03 m):
Work = F * d
Setting these two equal, we can solve for the average force exerted by the board:
F * d = - (1/2) * 0.025 kg * (300 m/s)^2
F = (- (1/2) * 0.025 kg * (300 m/s)^2) / 0.03 m
Calculating this expression, we find:
F = 37,500 N
Therefore, the average force exerted on the bullet by the hard board is 37,500 N.
B) To find the average force exerted on the bullet by the soft board in scenario B where the bullet emerges with a speed of 50 m/s, we can again use the principle of work and energy.
The work done on an object is equal to the product of the force applied and the distance it moves in the direction of the force. In this case, the force exerted by the board decelerates the bullet.
The initial kinetic energy of the bullet is given by:
KE_initial = (1/2) * m * v_initial^2
where m is the mass of the bullet (0.025 kg) and v_initial is the initial velocity of the bullet (300 m/s).
The final kinetic energy of the bullet when it emerges with a speed of 50 m/s is given by:
KE_final = (1/2) * m * v_final^2
where v_final is the final velocity of the bullet (50 m/s).
The work done by the board is equal to the change in kinetic energy of the bullet:
Work = KE_final - KE_initial
= (1/2) * m * v_final^2 - (1/2) * m * v_initial^2
Now, the work done by the board is equal to the force exerted by the board (F) multiplied by the distance through which the bullet moves in the board (d = 4.0 cm = 0.04 m):
Work = F * d
Setting these two equal, we can solve for the average force exerted by the board:
F * d = (1/2) * m * v_final^2 - (1/2) * m * v_initial^2
F = [(1/2) * m * v_final^2 - (1/2) * m * v_initial^2] / d
Calculating this expression, we find:
F = [(1/2) * 0.025 kg * (50 m/s)^2 - (1/2) * 0.025 kg * (300 m/s)^2] / 0.04 m
F = 27344 N
Therefore, the average force exerted on the bullet by the soft board is 27344 N.