Question

A police van moving on a highway with a speed of $\text{30 km}{{\text{h}}^{-\text{1}}}$ fires a bullet at a thief's car speeding away in the same direction with a speed of $\text{192 km}{{\text{h}}^{-\text{1}}}$. If the muzzle speed of the bullet is $\text{150 m}{{\text{s}}^{-\text{1}}}$, with what speed does the bullet hit the thief's car?
(A). $\text{95 m}{{\text{s}}^{-\text{1}}}$
(B). $\text{105 m}{{\text{s}}^{-\text{1}}}$
(C). $\text{115 m}{{\text{s}}^{-\text{1}}}$
(D). $\text{125 m}{{\text{s}}^{-\text{1}}}$

Answer 1

Hint: We can easily solve this question with the concept of relative motion. The relative velocity of the bodies will help determine the final velocity of any body as is shown in the solution below.

Complete Step-by-step solution:
From, the given question, we can determine that a police van moving on a highway has a speed of
${{\text{v}}_{\text{p}}}\text{ = 30 km}{{\text{h}}^{-1}}$
$=\text{ 30 }\times \text{ }\dfrac{1000\text{m}}{3600s}$
$=\text{ }\dfrac{25}{3}\text{ }{\text{m}}/{\text{s}}\;$
Similarly, we can determine that the speed of the thief’s car in the same direction is
${{\text{v}}_{\text{t}}}\text{ = 190 km}{{\text{h}}^{-1}}$
$=\text{ 190 }\times \text{ }\dfrac{1000\text{m}}{3600s}$
$=\text{ }\dfrac{160}{3}\text{ }{\text{m}}/{\text{s}}\;$
Similarly, we can determine that the bullet fired by the police in the same direction had a muzzle velocity of
$=\text{ 150 }{\text{m}}/{\text{s}}\;$
Since,
The direction of bullet fired and the police car is same, we can find the final velocity of the bullet fired from the gun as,
${{\text{v}}_{\text{b}}}\text{ + }{{\text{v}}_{\text{p}}}\text{ = }\left( \text{150 + }\dfrac{25}{3} \right){\text{m}}/{\text{s}}\;$
$=\text{ }\dfrac{475}{3}\text{ }{\text{m}}/{\text{s}}\;$
If we assume the velocity of the thief to be zero, with respect to the thief, we can determine the velocity of the bullet to be
${{\text{v}}_{{\text{b}}/{\text{t}}\;}}\text{ = }{{\text{v}}_{\text{b}}}\text{ }-\text{ }{{\text{v}}_{\text{t}}}$
$=\text{ }\dfrac{475}{3}\text{ }-\text{ }\dfrac{160}{3}$
$=\text{ }\dfrac{315}{3}\text{ }{\text{m}}/{\text{s}}\;$
$=\text{ 105 }{\text{m}}/{\text{s}}\;$
Therefore, we can see that the final velocity of the bullet with which it hits the thief’s car is $\text{105 }{\text{m}}/{\text{s}}\;$.
Hence, the correct option is Option B.

Note: In this question, we have seen the mentioning of a concept, which is muzzle speed. By muzzle speed we mean the speed of a projectile taken in accordance with the muzzle at the moment we are taken into consideration. By the considered time, we mean the time when the bullet will be leaving the end of a gun’s barrel.