Question

AbhiJeet fires a bullet of mass 100g with a speed of 100m/s on a soft plywood of thickness 4cm. The bullet emerges with 10% of its initial K.E. Find the emergent speed of bullet

Answer 1

Answer:

10√10 m/sec

Explanation:

(ki-kf) /ki 100% = (100-10)%

vi^2-vf^2/vi^2 100% = 90%

vi^2 - vf^2 = (9/10)vi^2

- vf^2 = -(1/10)vi^2

vf^2 = (1/10)(10000)

vf = 10√10 m/sec.

the bullet emerges with the speed of 10√10 m/sec.

Answer 2

Answer:

31.6ms-1

Explanation:

initial KE of bullet =1/2mu2 =1/2×100/1000×100×100= 500J

Final KE of the bullet =10/100×500=50J

Let emergent speed of the bullet be v, 1/2mv2 =50

v= √2×50/100/1000= 31.6ms-1