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PhysicsKinematicsMiscellaneousEasy2 minQB

PhysicsEasysingle choice

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.

z-coordinate of the point where ball lands, is-

Options:

Answer:

Solution:

Time of flight $t_{2} = \frac{2 \times 10}{10} = 2 s, z$ -coordinate $z = u_{z} (t_{1} + t_{2}) = 5 (3 + 2) = 25 m$

Ball walls projected at $t = 3 sec$

So ball will land after 3 + 2

i.e. 5sec So z-coordinate : $z = u_{z} t$

$\Rightarrow z = 5 \times (3 + 2) \Rightarrow z = 5 \times 5 \Rightarrow z = 25 m$

Hence, options (c) is correct.

Stream:JEESubject:PhysicsTopic:KinematicsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: QB

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