Question

A ball of mass 100g is projected vertically upwards from the ground with a velocity of 49m/s at the same time another identical ba is dropped from a height of 98m to fall freely alomg the same path as followed by the first ball. After sometime te two balls collide and stick together.

Answer 1

Answer:

Total time of the motion is T = 6.5 s

Explanation:

Since two balls start the motion at same time

So here we can use the relative motion to find the time after which they collide

so we have

relative velocity of two balls is v = 49 m/s

distance between two balls = 98 m

relative acceleration = g - g = 0

so the time of collision between two balls is given as

$t = \frac{d}{v}$

$t = \frac{98}{49} = 2 s$

now the distance at which they collide is given as

$d = H - \frac{1}{2}gt^2$

$d = 98 - \frac{1}{2}(9.8)(2^2)$

$d = 78.4 m$

now velocity of two balls just before collision

for ball which is dropped we have

$v_1 = (9.8)(2) = 19.6 m/s$ downwards

for ball which is projected upwards

$v_2 = 49 - (9.8)(2) = 29.4 m/s$ upwards

now we have

$m_1v_1 - m_2v_2 = (m_1 + m_2) v$

$100 (29.4 - 19.6) = (100 + 100) v$

$v = 4.9 m/s$

now time to hit the ground is given as

$y = vt + \frac{1}{2}at^2$

$-78.4 = 4.9 t - \frac{1}{2}(9.8) t^2$

$t^2 - t - 16 = 0$

$t = 4.5 s$

So total time of the motion is given as

$T = 4.5 + 2$

$T = 6.5 s$

#Learn

Topic : Momentum conservation

https://brainly.in/question/11838764

Answer 2

Answer:

Solution ⇒ Mass = 100 g. = 0.1 kg.

Speed (u) = 49 m/s.

θ = 45°

∵ H = u²Sin²θ/2g

⇒ H = (49)² × (1/√2)²/20

⇒ H = 60.025 m.

Now, T = 2u Sinθ/g

⇒ T = 98/10√2

⇒ T = 0.82 seconds.

Hope it helps.