Complete question is;
A 13g bullet traveling 230 m/s penetrates a 2.0 kg block of wood and emerges going 170 m/s. If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?
Answer:
0.39 m/s
Step-by-step explanation:
From the question, we can deduce that there is no external force acting on the
system in the horizontal direction. This means that momentum will be conserved for the system of the 2 kg block & the bullet.
Thus;
Initial momentum(p_i) = final momentum(p_f)
Mass of bullet; = 13g = 0.013 kg
Mass of Block = 2 kg
Speed of bullet = 230 m/s
Speed of Block = 170 m/s
We know that formula for momentum is; p = mv.
Thus, the initial momentum of the (bullet + block) is given as;
p_i = 0.013 × 230
Final momentum is;
p_f = (0.013 × 170) + (2 × v_f)
Where v_f is the speed at which the bullet emerges.
Since p_i = p_f from conservation of momentum, then;
(0.013 × 230) = (0.013 × 170) + (2 × v_f)
2.99 = 2.21 + 2v_f
2.99 - 2.21 = 2v_f
0.78 = 2v_f
v_f = 0.78/2
v_f = 0.39 m/s