Solveeit Logo
Login

Question

Question: In a ballistics demonstration a police officer fires a bullet of mass 50 g with speed \(200 \mathrm...

In a ballistics demonstration a police officer fires a bullet of mass 50 g with speed 200 ms−1200 \mathrm {~ms} ^ { - 1 } on soft plywood of thickness 2 cm. The bullet emerges with only 10% of its initial kinetic energy. The emergent speed of the bullet is :

A

210 ms−12 \sqrt { 10 \mathrm {~ms} ^ { - 1 } }

B

2010 ms−120 \sqrt { 10 \mathrm {~ms} ^ { - 1 } }

C

102 ms−110 \sqrt { 2 \mathrm {~ms} ^ { - 1 } }

D

1020 ms−110 \sqrt { 20 \mathrm {~ms} ^ { - 1 } }

Answer

2010 ms−120 \sqrt { 10 \mathrm {~ms} ^ { - 1 } }

Explanation

Solution

Here, m=50 g×10−3 kg\mathrm { m } = 50 \mathrm {~g} \times 10 ^ { - 3 } \mathrm {~kg}

Initial kinetic energy of the bullet is

Final kinetic energy of the bullet is

kf=10%ki=10100×1000 J=100 J\mathrm { k } _ { \mathrm { f } } = 10 \% \mathrm { k } _ { \mathrm { i } } = \frac { 10 } { 100 } \times 1000 \mathrm {~J} = 100 \mathrm {~J}

Let be the emergent speed of the bullet. Then

=10025 ms−1=2010 m s−1= 100 \sqrt { \frac { 2 } { 5 } } \mathrm {~ms} ^ { - 1 } = 20 \sqrt { 10 } \mathrm {~m} \mathrm {~s} ^ { - 1 }