Search any question & find its solution
Question:
Answered & Verified by Expert
A cricket ball of mass $150 \mathrm{~g}$ has an initial velocity $\vec{u}=(3 \hat{i}+4 \hat{j}) \mathrm{ms}^{-1}$ and a final velocity $\vec{v}=-(3 \hat{i}+4 \hat{j}) \mathrm{ms}^{-1}$, after being hit. The change in momentum (final momentuminitial momentum) is (in $\mathrm{kgms}^{-1}$ )
Options:
Solution:
2631 Upvotes
Verified Answer
The correct answer is:
$-(0.9 \hat{j}+1.2 \hat{j})$
$-(0.9 \hat{j}+1.2 \hat{j})$
As given that,
Mass of the ball $=150 \mathrm{~g}=0.15 \mathrm{~kg}$
$$
\begin{aligned}
&\vec{u}=(3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s} \\
&\vec{v}=-(3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}
\end{aligned}
$$
$(\Delta p)$ Change in momentum
$=$ Final momentum-Initial momentum
$=m \vec{v}-m \vec{u}$
$=m(\vec{v}-\vec{u})=(0.15)[-(3 \hat{i}+4 \hat{j})-(3 \hat{i}+4 \hat{j})]$
$=(0.15)[-6 \hat{i}-8 \hat{j}]$
$=-[0.15 \times 6 \hat{i}+0.15 \times 8 \hat{j}]$
$=-[0.9 \hat{i}+1.20 \hat{j}]$
$\Delta p=-[0.9 \hat{i}+1.2 \hat{j}]$
Hence verifies option (c).
Mass of the ball $=150 \mathrm{~g}=0.15 \mathrm{~kg}$
$$
\begin{aligned}
&\vec{u}=(3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s} \\
&\vec{v}=-(3 \hat{i}+4 \hat{j}) \mathrm{m} / \mathrm{s}
\end{aligned}
$$
$(\Delta p)$ Change in momentum
$=$ Final momentum-Initial momentum
$=m \vec{v}-m \vec{u}$
$=m(\vec{v}-\vec{u})=(0.15)[-(3 \hat{i}+4 \hat{j})-(3 \hat{i}+4 \hat{j})]$
$=(0.15)[-6 \hat{i}-8 \hat{j}]$
$=-[0.15 \times 6 \hat{i}+0.15 \times 8 \hat{j}]$
$=-[0.9 \hat{i}+1.20 \hat{j}]$
$\Delta p=-[0.9 \hat{i}+1.2 \hat{j}]$
Hence verifies option (c).
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.