[a bullet weighing 50gm hits a 10cm
wall with velocity 100ms-1 and comes out of the wall. If the frictional force on the bullet
while in the wall is assumed to be a constant 2000N then with what velocity does the bullet
come out of the wall?
To determine the velocity at which the bullet comes out of the wall, we can use the principle of conservation of momentum.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant before and after an event, as long as no external forces act on the system.
In this case, we can consider the bullet and the wall as an isolated system. Before the bullet hits the wall, it is moving with a velocity of 100 m/s. Let's assume that the bullet comes to rest while in the wall for a brief moment, and then comes out of the wall with a velocity of v m/s.
Using the conservation of momentum:
Initial momentum = Final momentum
The initial momentum of the system is the momentum of the bullet before it hits the wall, given by:
Momentumprevious = (mass of the bullet) x (initial velocity of the bullet)
The final momentum of the system is the sum of the momentum of the bullet after it comes out of the wall and the momentum acquired by the wall:
Momentumfinal = (mass of the bullet) x (velocity of the bullet after coming out of the wall) + (mass of the wall) x (velocity of the wall after the collision)
Since the wall is stationary and does not move, the velocity of the wall after the collision is 0 m/s.
Now let's plug in the numbers:
Momentumprevious = (0.050 kg) x (100 m/s)
Momentumfinal = (0.050 kg) x (v m/s) + 0 kg x (0 m/s)
Equating these two momenta:
(0.050 kg) x (100 m/s) = (0.050 kg) x (v m/s)
Solving for v, we get:
v = (0.050 kg) x (100 m/s) / (0.050 kg)
v = 100 m/s
Therefore, the bullet comes out of the wall with a velocity of 100 m/s.