bullet and a swinging sack of sand

Question

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.

Answer 1

This problem requires the use of the conservation of momentum law in a two-dimensional inelastic collision.

Let us label m = 0.01 kg and M = 9.99 kg (the mass of the bullet and sack, respectively).

Also, let v_1x = 200 m/s, v_1y = 0 m/s be the velocity of the bullet along the x and y direction before the collision (this is what [200, 0]m/s represents).

Similarly, the velocity of the sack before the collision is v_2x = 0 m/s, v_2y = 0.2 m/s.

Momentum must be conserved in this collision BOTH along the x and the y direction.

Along x: mv_1x + Mv_2x = (m+M)V_x

Along y: mv_1y + Mv_2y = (m+M)V_y

In these two equations, V_x and V_y represent the velocity of the bullet/sack system after the collision.

Plugging in numbers, we obtain:

0.01 x 200 + 9.99 x 0 = (0.01 + 9.99)V_x

0.01 x 0 + 9.99 x 0.2 = (0.01 + 9.99)V_y

V_x = 0.2 m/s, V_y = 0.1998 m/s

To find the final velocity, we add these two velocities keeping in mind that they are vectors. We use the Pythagorean Theorem:

V² = V_x² + V_y², V = 0.283 m/s