A bullet of mass 50g is fired with a velocity of (500,0)m/s in to a sack of sand of mass 20kg which is swinging from rope. At the moment the bullet hits, the sack has a velocity of (0,3)m/s. Workout the velocity of bullet and sack just after the bullet hits the sack.
Given: M1 = 0.05kg, V1= 500m/s[0o].
M2 = 20 kg, V2 = 3m/s[90o]. = 3i m/s.
V3 = velocity of M1 and M2 after collision.
Momentum before = Momentum after
M1*V1+M2*V2 = M1*V3+M2*V3
0.05*500+20*3i = 0.05V3+20*V3
25 + 60i = 20.05V3
V3 = 1.25 + 2.99i = 3.24m/s[67.3o].
To work out the velocity of the bullet and sack just after the bullet hits the sack, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Before the collision:
The momentum of the bullet before the collision is calculated by multiplying its mass (50g) by its velocity (500 m/s):
Momentum of bullet before = (50g) * (500,0)m/s = 25000 g*m/s
The momentum of the sack before the collision is calculated by multiplying its mass (20kg) by its velocity (0,3)m/s:
Momentum of sack before = (20kg) * (0,3)m/s = 6 kg*m/s
The total momentum before the collision is the vector sum of the individual momenta of the bullet and the sack:
Total momentum before = (25000 g*m/s) + (6 kg*m/s)
After the collision:
Since the bullet and the sack are a combined system after the collision, we need to find the velocity of the entire system.
Let the velocity of the system after the collision be (Vx, Vy) m/s.
The momentum of the bullet and the sack after the collision is given by their combined mass and velocity:
Momentum of bullet and sack after = ((50g + 20kg) * (Vx, Vy)m/s
According to the conservation of momentum, we can equate the total momentum before the collision to the momentum after the collision:
Total momentum before = Momentum of bullet and sack after
The x-component and y-component of momentum are separately conserved, so we can equate them separately:
25000 g*m/s = (50g + 20kg) * (Vx)m/s -- for the x-component
0 kg*m/s = (50g + 20kg) * (Vy)m/s -- for the y-component
Simplifying these equations, we get:
(50g + 20kg) * (Vx)m/s = 25000 g*m/s
(50g + 20kg) * (Vy)m/s = 0 kg*m/s
Now we can solve for Vx and Vy:
Vx = (25000 g*m/s) / (50g + 20kg)
Vy = 0 (since the y-component of momentum is conserved as 0)
Converting g to kg:
Vx = (25000 * 0.001kg*m/s) / (0.05kg + 20kg)
Vx = (25 kg*m/s) / (20.05 kg)
Vx = 1.25 m/s
Therefore, the velocity of the bullet and sack just after the bullet hits the sack is approximately (1.25, 0)m/s.