A bullet of mass 0.01kg is fired with velocity of [200,0]m/s in to a sack of sand of mass 9.99kg which is swinging from rope. At the moment the bullet hits, the sack has a velocity of [0,0.2]m/s. Workout the velocity of bullet and sand just after the bullet hits the sack
The velocity of the bullet and the sand just after the bullet hits the sack can be calculated using the conservation of momentum.
Bullet velocity: [200, 0] m/s
Sack velocity: [0, 0.2] m/s
Total momentum before collision: 200 kg m/s
Total momentum after collision: 200 kg m/s
Therefore, the velocity of the bullet and the sand just after the collision can be calculated as follows:
Bullet velocity: [200 - 0.2, 0] m/s = [199.8, 0] m/s
Sack velocity: [0 + 0.01, 0.2 + 0] m/s = [0.01, 0.2] m/s
Nonsense. The bullet is buried in the sand.
You have one object, (sand with buried bullet) moving with the original total momentum
original x momentum = 0.01 * 200 = 2 kg m/s
original y momentum = 9.99 * 0.2 = 1.998 kg m/s
FINAL mass = 0.01 + 9.99 = 2 kg
final u = 2/2 = 1 m/s
final v = 1.998 / 2 = 1 m/s
( 1 , 1 )