Question

Ball $1$ is released from the top of a smooth inclined plane, and at the same instant ball $2$ is projected from the foot of the plane with such a velocity that they meet halfway up the incline. Determine:

(a) the velocity with which balls are projected and

(b) the velocity of each ball when they meet.

Question Image

(a) the velocity with which balls are projected and

(b) the velocity of each ball when they meet.

Accepted Answer

For ball $1$ : $u_{1} = 0, a = g \sin θ = g (\frac{h}{l})$ and $s = \frac{l}{2}$

Using $v^{2} = u^{2} + 2 a s \Rightarrow v^{2} = 0 + 2 g (\frac{h}{l}) \cdot (\frac{l}{2})$

$\Rightarrow v = \sqrt{g h} . . . (i)$

Again using, $v = u + a t \Rightarrow v = 0 + g (\frac{h}{l}) \cdot t$
$\Rightarrow \sqrt{g h} = g (\frac{h}{l}) t$ [From $(i)$ ]

$\Rightarrow t = \frac{l}{\sqrt{g h}}$

For ball $2$ : $u = u_{2}, v = v_{2}$ and $a = - g \sin θ = - g (\frac{h}{l})$
Using $v^{2} = u^{2} + 2 a s \Rightarrow v_{2}^{2} = u_{2}^{2} + 2 (- g \frac{h}{l}) \cdot (\frac{l}{2}) . . . (iii)$

Also time taken by the second ball will be same as time taken by first ball to reach the half of the inclined.
Again using $v = u + a t \Rightarrow v_{2} = u_{2} - (g \frac{h}{l}) \cdot \frac{l}{\sqrt{g h}} . . . . (iv)$
From $(iii)$ and $(iv)$ , we get $u_{2} = \sqrt{g h}$ and $v_{2} = 0$
Hence, velocity of ball $2$ at the time of throwing will be $\sqrt{g h}$ and when it meets ball $1$ , its velocity becomes zero. The velocity of ball $1$ when it meets ball $2$ is $\sqrt{g h}$ .

B M Sharma Solutions for Chapter: Kinematics I, Exercise 35: ILLUSTRATION 4.33

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Kinematics I, Exercise 35: ILLUSTRATION 4.33

Questions from B M Sharma Solutions for Chapter: Kinematics I, Exercise 35: ILLUSTRATION 4.33 with Hints & Solutions

Learn and Prepare for any exam you want !