Question

A 10.4-g bullet is fired into a stationary block of wood having mass m = 4.99 kg. The bullet imbeds into the block. The speed of the bullet-plus-wood combination immediately after the collision is 0.592 m/s. What was the original speed of the bullet? (Express your answer with four significant figures.)

Answer 1

V_b=Speed of bullet

M_b=mass of bullet

V_w=speed of wood

M_w=mass of wood

conservation of momentum

M_b*V_b=M_w*V_w

0.0104*V_B=(4.99+0.0104)*0.592

V_b=284.64 m/s

Answer 2

Using momentum conservation principle;
10.4 *v + 0 *4990 = 0.592 *(10.4 + 4900)

we get;
v= 279.51 m/s

In four significant digits:

Answer : 279.5 m/s

Answer 3

Using momentum conservation principle;
10.4 *v + 0 *4990 = 0.592 *(10.4 + 4990)

we get;
v= 284.6 m/s