Question

Q. 13. A bullet of mass 20 g moving with a speed of 500 ms 1 strikes a wooden block of
mass 1 kg and gets embedded in it. Find the speed with which block moves alongwith
the bullet.
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Answer 1

Answer:

Explanation:

M1U1 + M2U2 = (M1 +M2) V

20/1000 × 500 + 1 × 0 = (20/1000 + 1) V

0.02 × 500 = (0.02 + 1) V

10 = 1.02 V

V=10/1.02

V=9.8 m/s

Answer 2

Answer:

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.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).

Explanation:

Hope it helps you...