Answer to Question #212364 in Classical Mechanics for bishal

Question

1.            A bullet moving at the rate of 300 m/s is fired into a thick target and penetrates up to 500 mm. If it is fired into a 250 mm thick target, find the velocity of emergence. Take the resistance to be uniform in both the cases.

Answer 1

The given information is

• The initial velocity is $V_{(o)}=300\;\text{m}/\text{s}$
• The initial distance traveled is $d_{(i)}=500\;\text{mm}$
• The final distance traveled is $d_{(f)}=250\;\text{mm}$

Converting the distances to meters.

Equivalence millimeters to meters. $1\;\text{m}=1000\;\text{mm}$

Multiplying by the conversion factor.

$d_{(i)}=500\;\text{mm}\times \dfrac{1\;\text{m}}{1000\;\text{mm}}=0.5\;\text{m}\\ d_{(f)}=250\;\text{mm}\times \dfrac{1\;\text{m}}{1000\;\text{mm}}=0.25\;\text{m}\\$

The acceleration that the bullet undergoes in the first impact is

$a=\dfrac{V_{(f)}^{2}-V_{(i)}^{2}}{2\;d_{(i)}}$

In which.

$V_{(i)}$ is the initial velocity.

$V_{(f)}$ is the final velocity.

$d_{(i)}$ is the distance

Evaluating numerically.

$a=\dfrac{V_{(f)}^{2}-V_{(i)}^{2}}{2\;d_{(i)}}\\ a=\dfrac{(0\;\text{m}/\text{s})^{2}-(300\;\text{m}/\text{s}^{2}}{2\times 0.5\;\text{m}}\\ a=-90,000\;\text{m}/\text{s}^{2}$

Now, for the second shot, the final velocity is given by

$V_{(f)}^{2}=V_{(i)}^{2}+2\;a\;d_{(f)}$

\\

Where.

In which.

$V_{(i)}$ is the initial velocity.

$V_{(f)}$ is the final velocity.

$d_{(f)}$ is the distance

Evaluating numerically.

$V_{(f)}^{2}=V_{(i)}^{2}+2\;a\;d_{(f)}\\ V_{(f)}=\sqrt{V_{(i)}^{2}+2\;a\;d_{(f)}}\\ V_{(f)}=\sqrt{(300\;\text{m}/\text{s})^{2}+2\times -90,000;\text{m}/\text{s}^{2}\times 0.25\;\text{m}}\\ V_{(f)}=\sqrt{ 45,000\text{m}^{2}/\text{s}^{2}}\\$

$V_{(f)}=212\;\text{m}/\text{s}$

Answer.

The velocity with which it emerges from the second target is $\displaystyle \color{red}{\boxed{V_{(f)}=212\;\text{m}/\text{s}}}$