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
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
2 answers
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 )
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 )
11 answers