# A 140 gm baseball with a velocity of 25.0 m/s is hit by a baseball bat and leaves at 30.0 m/s in the opposite direction. If the ball was in contact with the bat for 12.0 ms, what is the average force on the ball?   Option 1) 750 N Option 2) 642 N Option 3) 550 N Option 4) 482 N

As we discussed in concept

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

$\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \frac{1}{2}m_{1}v_{1}^{2}+\frac{1}{2}m_{2}v_{2}^{2}$

$m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}$

$m_{1},m_{2}:masses$

$u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}$

$u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}$

$\vec{F}=\frac{\vec{\Delta P}}{t}=\frac{m(25-(-30))}{12\times 10^{-3}}$

$\vec{F}=\frac{140\times10^{-3}\times 55}{12\times 10^{-3}}\:=\:\frac{140\times 55}{12}\:=\:641.6=642N$

Option 1)

750 N

This option is incorrect.

Option 2)

642 N

This option is correct.

Option 3)

550 N

This option is incorrect.

Option 4)

482 N

This option is incorrect.

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