Explanation:
$\text{W}=\text{Fs}=\cos\theta=0$, when either s = 0 or $\theta=90^\circ$ i.e., when object is stationary but the point of application of the force moves on the object or object moves in such a way that point of application of force remains fixed; or force is at 90° to the acceleration.
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Figure: $Image$
$1.$ The speed of the block at point $B$ immediately after it strikes the second incline is
$(A)$ $\sqrt{60} \mathrm{~m} / \mathrm{s}$ $(B)$ $\sqrt{45} \mathrm{~m} / \mathrm{s}$
$(C)$ $\sqrt{30} \mathrm{~m} / \mathrm{s}$ $(D)$ $\sqrt{15} \mathrm{~m} / \mathrm{s}$
$2.$ The speed of the block at point $\mathrm{C}$, immediately before it leaves the second incline is
$(A)$ $\sqrt{120} \mathrm{~m} / \mathrm{s}$ $(B)$ $\sqrt{105} \mathrm{~m} / \mathrm{s}$
$(C)$ $\sqrt{90} \mathrm{~m} / \mathrm{s}$ $(D)$ $\sqrt{75} \mathrm{~m} / \mathrm{s}$
$3.$ If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point $B$, immediately after it strikes the second incline is
$(A)$ $\sqrt{30} \mathrm{~m} / \mathrm{s}$ $(B)$ $\sqrt{15} \mathrm{~m} / \mathrm{s}$
$(C)$ 0 $(D)$ $-\sqrt{15} \mathrm{~m} / \mathrm{s}$
Give the answer question $1,2$ and $3.$
$Reason$ : $\frac{{\Delta E}}{E} = \frac{{\Delta m}}{m} + \frac{{2\Delta v}}{v}$
