Question

A marble moving with some velocity collides perfectly elastically head-on with another marble at rest having mass 1.5 times the mass of the colliding marble . The percentage of kinetic energy by the colliding marble after the collision is
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Answer 1

Answer:

A marble moving with some velocity collides perfectly elastically head-on with another marble at rest having mass 1.5 times the mass of the colliding marble. The percentage of kinetic energy by the colliding marble after the collision is $4$%    

Explanation:

A marble collides another marble at rest elastically, since the collision is head-on initial and final velocities of both the bodies are along the same straight line as shown in the figure.

Let the mass of the first marble be $m_{1}$ and that of the second marble $m_{2} =1.5 m_{1}$

Assume the velocity of the first marble is $u_{1}$ before the collision, and $v_{1}$, $v_{2}$ are the velocities of the first and second marble after the collision.

For an elastic collision, both momentum and kinetic energy are conserved.

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

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

After rearranging the terms we get,

$v_{1}=(\frac{m_{1} -m_{2} }{m_{1} +m_{2} })u_{1}$

The percentage of the kinetic energy of the colliding marble is,

$(\frac{1}{2}m_{1}u_{1} ^{2}/\frac{1}{2}m_{1}v_{1} ^{2})100$

Cancel similar terms and substitute the value of masses

$\frac{u_{1} }{v_{1} }=(\ \frac{m_{1} -1.5m _{1} }{m_{1} +1.5m_{1} })$$=1/5$

$\frac{u_{1} ^{2} }{v_{1} ^{2} } =1/25=0.04$

The percentage of the kinetic energy of the colliding marble is,

$0.04*100=4$%