Question

"A bullet of mass 0.01 kg and travelling at a speed of 500m/s strikes a block which is suspended by a string of length 5m. The centre of gravity of the block is found to rise a distance of 0.1m. What is the speed of bullet after it emerges from the block?"

Answer 1

by energy conservation of the suspended block we can say

initial KE = final potential energy

$\frac{1}{2}mv^2 = mgH$

$\frac{1}{2}v^2 = gH$

$v = \sqrt{2gH}$

$v = \sqrt{2*10*0.1} = 1.41 m/s$

now by momentum conservation

$P_{bullet} = P_{block} + P_{bullet}'$

$0.01* 500 = 1* 1.41 + 0.01 * v_f$

$5 - 1.41 = 0.01*v_f$

$0.01v_f = 3.59$

$v_f = 359 m/s$

so finally bullet will move out of the block with speed 359 m/s

Answer 2

Answer:

by energy conservation of the suspended block we can say

initial KE = final potential energy

1/2 mv^2 = mgH

1/2 v^2 = gH

v = root(2gh)

v = root(2*10*0.1) = 1.41 m/s

now by momentum conservation

P_{bullet} = P_{block} + P_{bullet}'

0.01 * 500 = 1.41 * 2 + 0.01 * v(f)

5 = 2.82 + vf/100

2.18 = vf / 100

vf = 220(approx)

Please mark as brainliest