Question

Two carts on an air track undergo a head‐on perfectly elastic collision. The first cart, with 0.060 kg mass, is initially moving in the +x direction with speed 4.5 m/s. The second cart, with mass 0.090 kg is also moving in the +x direction, but with a speed of 3.0 m/s. Calculate the speed and direction of each cart following the collision.

Answer 1

The final speed of cart with mass 0.06 kg is 2.7 m/s and the final speed of the cart of mass 0.09 kg is 4.2 m/s.

Step-by-step explanation:

Initial moumentum of cart with mass 0.06 kg is

`p_1=0.06* 4.5=0.27kgm/s`

Initial moumentum of cart with mass 0.06 kg is

`p_2=0.09* 3=0.27kgm/s`

Thus initial moumentum of system is

`p_1+p_2=2* 0.27=0.54kgm/s`

Now let the final velocity of 0.06 kg cart be
`v_(f1)` and that of 0.09 kg cart be
`v_(f2)`

Thus final moumentum of system is

`0.06* v_(f1)+0.09* v_(f2)`

Equating initial and final values we get

`0.06* v_(f1)+0.09* v_(f2)=0.54........(i)`

Similarly conserving the kinetic energies of the 2 carts we get

`(1)/(2)* 0.06* 4.5^(2)+(1)/(2)* 0.09* 3^(2)=(1)/(2)* 0.06* (v_(1f))^(2)+(1)/(2)* 0.09* (v_(2f))^(2)\\\\0.06* 4.5^(2)+0.09* 3^(2)=0.06* (v_(2f))}^(2)+0.09* (v_(2f))^(2)\\\\0.06* (v_(2f))}^(2)+0.09* (v_(2f))^(2)=2.025.......(ii)`

Solving equation i and ii we get

Substituting value of
`v_(2f)` from i and using it in equation ii we get

`0.06v_(1f)^2+(0.06^2)/(0.09)v_(1f)^(2)-2* (0.54* 0.06)/(0.09)v_(1f)+(0.54^2)/(0.09)=2.025`

Solving for
`v_(1f)` we get

`v_(1f)=2.7m/s`

Similarly

`v_(2f)=(0.54-0.06* 2.7)/(0.09)=4.2m/s`