Question

In an ballistics demonstration a police officer fires a bullet of mass 50.0g with speed 200m/s on soft plywood of thickness 2.00cm.the bullet emerges with only 10% of its initial kinetic energy. what is the emergent speed of the bullet?​

Answer 1

Answer:

Hello mate,

Explanation:

Here, m=50.0g

=50/1000kg=1/20kg

vi=200ms−1

∴ Initial K.E., Ki=1/2mv²i

=1/2×1/20(200)²=1000J

Final K.E., Kf=10%(Ki)

=10/100×1000J=100J

If vf is emergent speed of the bullet, then

1/2mv²f=Kf

vf=√2Kf/m−−−−√=2×100/1/20−−−−−−−√=63.2m/s

Note that K.E. is reduced by 90% but speed is reduced by nearly 68%.

hope this helps u

Answer 2

$\large \underline{ \tt{Given : }} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{m = 50 g} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{u = 200 m/s} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ v = \: ?}$

$\tt{final \: \: k.e = \frac{1}{2} \: m {v}^{2} }$

$\tt{initial \: \: \: k.e \: \: \: = \: \frac{1}{2} \: m {u}^{2} }$

$\tt{ = \frac{1}{2} ( \frac{50}{1000} ) \times ({200})^{2} }$

$\tt{ = \frac{50}{2000} \times 40000}$

$\tt{ = 1000 \: \: joules}$

$\tt{final \: \: k.e = 10\% \: \: of \: \: initial \: \: k.e}$

$\tt{final \: \: k.e = \frac{10}{100} \times (1000)}$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \tt{ = 100 \: \: joules}$

$\tt{ \frac{1}{2} \: m {v}^{2} = 100}$

$\tt{ \frac{1}{2} ( \frac{50}{1000} ) {v}^{2} = 100}$

$\tt{ \frac{50}{2000} {v}^{2} = 100 }$

$\tt{50 {v}^{2} = 100 \times 2000 }$

$\tt{50 {v}^{2} = 200000}$

$\tt{ {v}^{2} = \frac{200000}{50} }$

$\tt{ {v}^{2} = \frac{20000}{5} }$

$\tt{ {v}^{2} = 4000}$

$\tt{v = \sqrt{4000} }$

$\large \boxed{\tt{v = 63.2 \: m/s }}$

$\\ \\ \\ \\ \red{ \bf{Hope \: \: this \: \: helps \: \: you \: !}}$