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A woman throws an object of mass $500 \mathrm{~g}$ with a speed of $25 \mathrm{~ms}^{-1}$.
(a) what is the impulse imparted to the object?
(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?
(a) what is the impulse imparted to the object?
(b) If the object hits a wall and rebounds with half the original speed, what is the change in momentum of the object?
Solution:
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Verified Answer
As given that,
Mass of the object $(\mathrm{m})=500 \mathrm{~g}=0.5 \mathrm{~kg}$
Speed of the object $(v)=25 \mathrm{~m} / \mathrm{s}, u=0$
(a) Impulse imparted to the object is equal to the change in momentum.
Impulse $=\vec{F} \cdot d t=$ change in momentum
$=m(\vec{v}-\vec{u})$
$$
d \vec{P}=0.5(25-0)=12.5 \mathrm{~N}-\mathrm{s}
$$
(b) Velocity of the object after rebounding
$$
\begin{aligned}
&=-\frac{25}{2} \mathrm{~m} / \mathrm{s} \text { (as backward) } \\
&v^{\prime}=-12.5 \mathrm{~m} / \mathrm{s} \\
&\therefore \Delta P(\text { Change in momentum })=m\left(v^{\prime}-v\right) \\
&\Delta P^{\prime}=0.5(-12.5-25)=-18.75 \mathrm{~N}-\mathrm{s}
\end{aligned}
$$
Hence, the $\Delta P$ or $\left(\frac{\Delta P}{\Delta t}\right)$ or force is opposite to the initial velocity of ball.
Mass of the object $(\mathrm{m})=500 \mathrm{~g}=0.5 \mathrm{~kg}$
Speed of the object $(v)=25 \mathrm{~m} / \mathrm{s}, u=0$
(a) Impulse imparted to the object is equal to the change in momentum.
Impulse $=\vec{F} \cdot d t=$ change in momentum
$=m(\vec{v}-\vec{u})$
$$
d \vec{P}=0.5(25-0)=12.5 \mathrm{~N}-\mathrm{s}
$$
(b) Velocity of the object after rebounding
$$
\begin{aligned}
&=-\frac{25}{2} \mathrm{~m} / \mathrm{s} \text { (as backward) } \\
&v^{\prime}=-12.5 \mathrm{~m} / \mathrm{s} \\
&\therefore \Delta P(\text { Change in momentum })=m\left(v^{\prime}-v\right) \\
&\Delta P^{\prime}=0.5(-12.5-25)=-18.75 \mathrm{~N}-\mathrm{s}
\end{aligned}
$$
Hence, the $\Delta P$ or $\left(\frac{\Delta P}{\Delta t}\right)$ or force is opposite to the initial velocity of ball.
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