A Small ball falling vertically downward with constant velocity $4 m / s$ strikes elastically a massive inclined cart moving with velocity $4 m / s$ horizontally as shown.The speed of the rebound of the ball is $4 \sqrt{P} m / s$ . Find value of $p$ .

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Solution

Ref image I
$\Rightarrow e = \frac{r e l a t i v e v e l o c i t y o f s e p e r a t i o n}{r e l a t i v e v e l o c i t y o f i m p a c t}$
$\Rightarrow 1 = \frac{4 sin 45 ° + 4 cos 45 °}{V - 4 sin 45 °}$
Let massive mass velocity doesn't change.
$\Rightarrow V - 2 \sqrt{2} = 4 \sqrt{2}$
$V = 6 \sqrt{2}$
Ref. image II
$\Rightarrow V_{0} = \sqrt{{(6 \sqrt{2})}^{2} + {(2 \sqrt{2})}^{2}}$
$= \sqrt{72 + 8}$
$= \sqrt{80}$
$= 4 \sqrt{5}$
$\Rightarrow p = 5.$
Hence, the answer is $5.$

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A Small ball falling vertically downward with constant velocity 4 m/s strikes elastically a massive inclined cart moving with velocity 4 m/s horizontally as shown.The speed of the rebound of the ball is 4√P m/s. Find value of p.

Ref image I⇒e=relativevelocityofseperationrelativevelocityofimpact⇒1=4sin45°+4cos45°V−4sin45°Let massive mass velocity doesn't change.⇒V−2√2=4√2 V=6√2Ref. image II⇒V0=√(6√2)2+(2√2)2 =√72+8 =√80 =4√5⇒p=5.Hence, the answer is 5.

A Small ball falling vertically downward with constant velocity $4 m / s$ strikes elastically a massive inclined cart moving with velocity $4 m / s$ horizontally as shown.The speed of the rebound of the ball is $4 \sqrt{P} m / s$ . Find value of $p$ .