- A$36$
- ✓$20$
- C$8$
- D$6$
In given problem $K.E.$ becomes $64\%$ of the original value.
$\frac{{{P_2}}}{{{P_1}}} = \sqrt {\frac{{{E_2}}}{{{E_1}}}} = \sqrt {\frac{{64E}}{{100E}}} = 0.8$==>${P_2} = 0.8\,P$
$\therefore {P_2} = 80\% $ of the original value.
i.e. decrease in momentum is $20\%$.
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$(A)$ the point $O$ has linear velocity $3 R \omega \hat{ i }$.
$(B)$ the point $P$ has a linear velocity $\frac{11}{4} R \omega \hat{ i }+\frac{\sqrt{3}}{4} R \omega \hat{ k }$
$(C)$ the point $P$ has linear velocity $\frac{13}{4} R w \hat{ i }-\frac{\sqrt{3}}{4} R \omega \hat{k}$
$(D)$ The point $P$ has a linear velocity $\left(3-\frac{\sqrt{3}}{4}\right) R w_{\hat{ i }}+\frac{1}{4} R w_{\hat{k}}$.
where $x=$ displacement at time $t$
$\omega =$ frequency of oscillation
Which one of the following graphs shows correctly the variation $a$ with $t$ ?
Here $a=$ acceleration at time $t$
$T=$ time period