A stunt man weighing 80kg drops himself gently into an open car weighing 720kg and moving with a velocity of 80kmph. The velocity of the car after the event is
(A) 36kmph
(B) 72 kmph
(C ) 80 kmph
(D) 78 kmph
Answer
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Hint: Here the net force is zero. Then by applying Newton’s second law of motion we get the force is equal to the rate of change of momentum. Thus the rate of change of momentum is zero here. Hence, the linear momentum will be conserved. Since the momentum can be described as the product of the mass and its velocity. Thus by substituting the initial mass and velocity and final mass we will get the final velocity.
Complete step-by-step solution
The net force acting along here is equal to zero. That is,
$\overrightarrow{{{F}_{net}}}=\overrightarrow{0}$ ………….(1)
According to Newton’s second law the force is equal to the rate of change of momentum. Mathematically this can be written as,
${{\overrightarrow{F}}_{net}}=\dfrac{\overrightarrow{dp}}{dt}$ ……….(2)
Equating equation(1) and (2) we get,
$\Rightarrow \dfrac{\overrightarrow{dp}}{dt}=\overrightarrow{0}$
$\Rightarrow d\overrightarrow{p}=0$
This shows that the initial momentum and final momentum are equal.
$\Rightarrow {{\overrightarrow{P}}_{i}}={{\overrightarrow{P}}_{f}}$
Thus, linear momentum will be conserved.
${{P}_{i}}={{P}_{f}}$
That is, the momentum is the product of the mass and its velocity.
$P=mv$
${{P}_{i}}={{P}_{f}}$
Since the momentum is conserved it can also be written as,
$\Rightarrow {{m}_{i}}{{v}_{i}}={{m}_{f}}{{v}_{f}}$ ………….(1)
Given that,
$\begin{align}
& {{m}_{i}}=720kg \\
& {{v}_{i}}=80kmph \\
& {{m}_{2}}=720+80=800kg \\
& \\
\end{align}$
Rearranging the equation (1) we get,
${{v}_{f}}=\dfrac{{{m}_{i}}{{v}_{i}}}{{{m}_{f}}}$
Substituting the values we get,
${{v}_{f}}=\dfrac{720\times 80}{800}$
$\therefore {{v}_{f}}=72kmph$
Thus option (B) is correct.
Note: According to Newton’s second law the force is equal to the rate of change of momentum. When the net force is zero, the linear momentum is conserved. That is the initial momentum and final momentum are equal here. And the momentum is the product of the mass and its velocity. Thus from the conservation of momentum product of initial mass and velocity will be equal to the product of final mass and velocity.
Complete step-by-step solution
The net force acting along here is equal to zero. That is,
$\overrightarrow{{{F}_{net}}}=\overrightarrow{0}$ ………….(1)
According to Newton’s second law the force is equal to the rate of change of momentum. Mathematically this can be written as,
${{\overrightarrow{F}}_{net}}=\dfrac{\overrightarrow{dp}}{dt}$ ……….(2)
Equating equation(1) and (2) we get,
$\Rightarrow \dfrac{\overrightarrow{dp}}{dt}=\overrightarrow{0}$
$\Rightarrow d\overrightarrow{p}=0$
This shows that the initial momentum and final momentum are equal.
$\Rightarrow {{\overrightarrow{P}}_{i}}={{\overrightarrow{P}}_{f}}$
Thus, linear momentum will be conserved.
${{P}_{i}}={{P}_{f}}$
That is, the momentum is the product of the mass and its velocity.
$P=mv$
${{P}_{i}}={{P}_{f}}$
Since the momentum is conserved it can also be written as,
$\Rightarrow {{m}_{i}}{{v}_{i}}={{m}_{f}}{{v}_{f}}$ ………….(1)
Given that,
$\begin{align}
& {{m}_{i}}=720kg \\
& {{v}_{i}}=80kmph \\
& {{m}_{2}}=720+80=800kg \\
& \\
\end{align}$
Rearranging the equation (1) we get,
${{v}_{f}}=\dfrac{{{m}_{i}}{{v}_{i}}}{{{m}_{f}}}$
Substituting the values we get,
${{v}_{f}}=\dfrac{720\times 80}{800}$
$\therefore {{v}_{f}}=72kmph$
Thus option (B) is correct.
Note: According to Newton’s second law the force is equal to the rate of change of momentum. When the net force is zero, the linear momentum is conserved. That is the initial momentum and final momentum are equal here. And the momentum is the product of the mass and its velocity. Thus from the conservation of momentum product of initial mass and velocity will be equal to the product of final mass and velocity.
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