Question

A bullet of mass 20 g moving with a speed of 500 m/s strikes a wooden block of mass 1 kg and gets embedded in it. Find the speed with which block moves along with the bullet.

Answer 1

Heya!!
Apply the principle of conservation of momentum here .

Mass of bullet = m = 20g or 20×10-³ Kg
Mass of block = M = 1 kg
Initial speed of bullet (v1) = 500m/s
final speed of bullet(v2) = 0
Initial speed of block (V1) = 0
Final speed of block (V2) = ❔

we know according to principal of conservation of linear momentum :- Initial momentum of the system = final momentum of the system .

so
M1V1 + m1v1 = M2V2 + m2v2
Now plug in the values , we get
1×0 + 20×10-³×500 = 1×V2 + 20×10^-³ ×0

=> V2 = 10000 ×10-³
=> V2 = 10m/s.

Answer 2

By applying law of conservation of momentum,

Combine velocity = m1u1 / (m1 + m2)
= (20 * 10^-3 * 500) / ((20 * 10^-3) + 1)
= 9.80 m/s
≈ 10 m/s

Speed with which block moves along with bullet is 9.80 m/s (approximately 10 m/s).