A block of metal weighing 2 kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1 kgs -1 and at a speed of 10 ms -1. Then, the initial acceleration of the block, in ms -2, will be :

Question

A block of metal weighing 2 kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1 kgs -1 and at a speed of 10 ms -1. Then, the initial acceleration of the block, in ms -2, will be : (A) 3 (B) 6 (C) 5 (D) 4

Tardigrade Admin · Accepted Answer

F =( dp / dt )=v ( dm / dt ) ⇒ Ma =10 × 1 ⇒ 2 a =10 a =5 m / sec 2

Q. A block of metal weighing $2 k g$ is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of $1 k g s^{- 1}$ and at a speed of $10 m s^{- 1}$ . Then, the initial acceleration of the block, in $m s^{- 2}$ , will be :

3

6

5

4

$F = \frac{d p}{d t} = v \frac{d m}{d t}$
$\Rightarrow M a = 10 \times 1$
$\Rightarrow 2 a = 10$
$a = 5 m / se c^{2}$

Q. A block of metal weighing 2kg is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of 1kgs−1 and at a speed of 10ms−1. Then, the initial acceleration of the block, in ms−2, will be :

3

6

5

4

F=dtdp​=vdtdm​ ⇒Ma=10×1 ⇒2a=10 a=5m/sec2

Q. A block of metal weighing $2 k g$ is resting on a frictionless plane (as shown in figure). It is struck by a jet releasing water at a rate of $1 k g s^{- 1}$ and at a speed of $10 m s^{- 1}$ . Then, the initial acceleration of the block, in $m s^{- 2}$ , will be :

$F = \frac{d p}{d t} = v \frac{d m}{d t}$
$\Rightarrow M a = 10 \times 1$
$\Rightarrow 2 a = 10$
$a = 5 m / se c^{2}$