Projection velocity of ball w.r.t. ground is | TestHub

Passage / Comprehension

The system shown in the figure is initially at origin and is moving with a velocity $(5 ms^{- 1}) \hat{k}$ . A force( 120 t newton) $\hat{i}$ acts on mass $m_{2}$ , where t is time in seconds. The man throws a ball of very small mass, at the instant when $m_{1}$ starts slipping on $m_{2}$ , with a velocity $10 ms^{- 1}$ vertically up w.r.t himself. Taking the masses of blocks and man as 60 kg each and assuming that the man never slips on $m_{2}$ . Take $g = 10 ms^{- 2}$ and neglect the dimensions of the system.

Projection velocity of ball w.r.t. ground is

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Answer:

Solution:

Projection velocity of ball is $u_{x} \hat{i} + u_{y} \hat{j} + u_{z} \hat{k}$ . Now, $u_{y} = 10 \hat{j}$ and $u_{z} = 5 \hat{k}$ . Since, in the x-direction, motion is accelerated:

$\frac{d u _{x}}{d t} = a_{x}$

$\int_{0}^{u_{x}} d v_{x} = \int a_{x} \cdot d t$

$u_{x} = \int a_{x} \cdot d t$

Now, from $0 < t \leq 3 s$ , the system will have the same acceleration. So, $F = (m_{1} + m_{2} + m_{3}) a_{x}$ .

$120 t = (180) a_{x}$

$a_{x} = \frac{2 t}{3}$

Now, $u_{x} = \int_{0}^{3} a_{x} \cdot d t$ .

$u_{x} = \int_{0}^{3} \frac{2 t}{3} \cdot d t = \frac{2}{3} [\frac{t ^{2}}{2}]_{0}^{3}$

$u_{x} = 3 m / s$

So, the velocity of projection is:

$3 \hat{i} + 10 \hat{j} + 5 \hat{k}$

Option (A) is correct.

Physics - Kinematics Question with Solution | TestHub

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