A bullet of mass0.01kg is fired with a velocity of 200m/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 0.2m/s. Workout the velocity of bullet and sand just after the bullet hits the sack.

You did not say if the sack was moving toward the rifle or away at start

initial momentum = 0.01 * 200 plus or minus 9.99 * 0.20
final momentum = (0.01 + 9.99) v
so
10 v = 2 + or - 2

Its correct or

To solve this problem, we can apply the law of conservation of momentum, which states that the total momentum before an event is equal to the total momentum after the event.

Before the bullet hits the sack, the total momentum is given by the sum of the momentum of the bullet and the momentum of the sack:

Total momentum before = (Mass of bullet × Velocity of bullet) + (Mass of sack × Velocity of sack before)

Total momentum before = (0.01 kg × 200 m/s) + (9.99 kg × 0.2 m/s)

Total momentum before = 2 kg·m/s + 1.998 kg·m/s

Total momentum before = 3.998 kg·m/s

After the bullet hits the sack, the total momentum is given by the sum of the final momentum of the bullet and the final momentum of the sack.

Let the final velocity of the bullet be v1 and the final velocity of the sack be v2.

Total momentum after = (Mass of bullet × Velocity of bullet after) + (Mass of sack × Velocity of sack after)

Total momentum after = (0.01 kg × v1) + (9.99 kg × v2)

According to the conservation of momentum, the total momentum before the event is equal to the total momentum after the event.

3.998 kg·m/s = (0.01 kg × v1) + (9.99 kg × v2)

We also know that the velocity of the sack before the bullet hits is 0.2 m/s. Therefore, the velocity of the sack after the bullet hits is also 0.2 m/s.

Plugging in these values, we can solve for v1:

3.998 kg·m/s = (0.01 kg × v1) + (9.99 kg × 0.2 m/s)

3.998 kg·m/s = 0.01 kg·v1 + 1.998 kg·m/s

0.01 kg·v1 = 3.998 kg·m/s - 1.998 kg·m/s

0.01 kg·v1 = 2 kg·m/s

v1 = (2 kg·m/s) / 0.01 kg

v1 = 200 m/s

Therefore, the final velocity of the bullet just after it hits the sack is 200 m/s.

To find the final velocity of the sand (sack), we can plug this result back into the conservation of momentum equation:

3.998 kg·m/s = (0.01 kg × 200 m/s) + (9.99 kg × v2)

3.998 kg·m/s = 2 kg·m/s + (9.99 kg × v2)

2 kg·m/s + (9.99 kg × v2) = 3.998 kg·m/s

(9.99 kg × v2) = 1.998 kg·m/s

v2 = (1.998 kg·m/s) / 9.99 kg

v2 = 0.2 m/s

Therefore, the final velocity of the sand just after the bullet hits the sack is 0.2 m/s.

To find the velocity of the bullet and sand just after the bullet hits the sack, we can apply the principle of conservation of momentum.

According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision. The momentum is given by the product of mass and velocity.

Before the collision, the bullet's momentum (p1) is given by:
p1 = m1 * v1
where m1 is the mass of the bullet and v1 is the velocity of the bullet.

After the collision, the momentum of the bullet (p2) and the sack (p3) are given by:
p2 = m1 * v2
p3 = m3 * v3
where m3 is the mass of the sack and v3 is the velocity of the sack.

Since the total momentum before the collision is equal to the total momentum after the collision, we can write:
p1 = p2 + p3

Substituting the values given in the problem:
0.01kg * 200m/s = 0.01kg * v2 + 9.99kg * 0.2m/s

Simplifying the equation:
2kgm/s = 0.01kg * v2 + 1.998kgm/s

Rearranging the equation to solve for v2:
0.01kg * v2 = 2kgm/s - 1.998kgm/s
0.01kg * v2 = 0.002kgm/s

Dividing both sides by 0.01kg:
v2 = 0.002kgm/s / 0.01kg
v2 = 0.2m/s

Therefore, the velocity of the bullet just after it hits the sack is 0.2m/s.

To find the velocity of the sand just after the bullet hits the sack, we substitute the value of v2 into the equation for p3:
p3 = m3 * v3
9.99kg * 0.2m/s = 1.998kgm/s

Therefore, the velocity of the sand just after the bullet hits the sack is 0.2m/s.