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Give answer! - Laws of motion - JEE Main

The figure shows the position - time (x-t)    graph of one-dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is

  • Option 1)

    0.2 N s

  • Option 2)

    0.4 N s

  • Option 3)

    0.8 N s

  • Option 4)

    1.6 N s

 
Answers (1)
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As we learnt in

Impulse Momentum Theorem -

\vec{F}=\frac{d\vec{p}}{dt}

\int_{t_{1}}^{t_{2}}\vec{F}dt=\int_{p_{1}}^{p_{2}}\vec{dp}

- wherein

If \bigtriangleup t  is increased, average force is decreased

 

 Initial velocity,  u=\frac{(2-0)}{2-0}=1ms^{-1}

Final velocity,  u=\frac{(0-2)}{4-2}=-1ms^{-1}

Initial Momentum P_{i}=mu=0.4\times 1=0.4\ N.s.

Final Momentum P_{f}=mv=0.4\times (-1)=-0.4\ N.s

Impulse = Change in Momentum = \overrightarrow{\Delta P}=\overrightarrow{P_{f}}=\overrightarrow{P_{i}}=-0.4-(0.4)N.s=-0.8\ N.s

\therefore\ \; \left | Impulse\right |=0.8\ N.s

Correct option is .3

 


Option 1)

0.2 N s

This is an incorrect option.

Option 2)

0.4 N s

This is an incorrect option.

Option 3)

0.8 N s

This is the correct option.

Option 4)

1.6 N s

This is an incorrect option.

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