Question

A ball of mass 50gm is dropped from a height \[h=10m\]. It rebounds losing \[75%\] of its kinetic energy. If it remains in contact with the ground for \[\Delta t=0.01\sec s\] , the impulse of the impact force will be-
(A). \[1.3N\,s\]
(B). \[1.05N\,s\]
(C). \[1300N\,s\]
(D).\[105N\,s\]

Answer 1

Hint: Impulse is product of force and change in time interval. It can also be calculated as change in momentum. The mechanical energy of the whole system is conserved, so the whole energy of the system converts to kinetic energy just before collision. The mechanical energy of the system decreases after collision as energy is lost.

Formula used: \[I=\Delta P\]
\[P=\sqrt{2mE}\]

Complete step by step answer:
Impulse is the product of force and change in time interval. It is given by-
\[I=F\Delta t\]
We know that the rate of change of momentum (\[P\]) is force. Therefore, from eq (1), we have,
\[I=\left( \dfrac{\Delta P}{\Delta t} \right)\Delta t\]
\[\therefore I=\Delta P\] ------- (1)
Just before collision, the kinetic energy of the ball will be maximum,
\[\begin{align}
  & E=50\times {{10}^{-3}}kg\times 10m\,{{s}^{-2}}\times 10 \\
 &\Rightarrow E=5J \\
\end{align}\]
Momentum just before collision of ball with the ground
\[\begin{align}
  & P=\sqrt{2mE} \\
 & \Rightarrow P=\sqrt{2\times 50\times {{10}^{-3}}\times 5} \\
\end{align}\]
\[\Rightarrow P=0.705kg\,m\,{{s}^{-1}}\] ----------- (2)
Kinetic energy decreases by\[75%\], so the new kinetic energy will be-
\[\begin{align}
  & \Delta E=0.75\times E \\
 &\Rightarrow \Delta E=0.75\times 5=3.75J \\
 &\Rightarrow E'=5-3.75=1.25J \\
\end{align}\]
Momentum just after collision with ground will be-
\[\begin{align}
  & P'=\sqrt{2mE'} \\
 &\Rightarrow P'=\sqrt{2\times 50\times {{10}^{-3}}\times 1.25} \\
\end{align}\]
\[\Rightarrow P'=0.353kg\,m\,{{s}^{-1}}\] -------- (3)
Calculating impulse from eq (1), eq (2), and eq (3), we get,
\[\begin{align}
  & I=P'-(-P) \\
 &\Rightarrow I=0.353+0.705=1.058\,N\,s \\
\end{align}\]

So, the correct answer is “Option B”.

Note: Momentum before and after collision are opposite in direction, that is why we will consider the final momentum as positive and initial momentum as negative. Impulse describes the overall effect of force in a given time interval. Impulse applied to an object changes its magnitude as well as direction.