Several students are riding bumper cars. Combined massof car A and its occupants

Question

Several students are riding bumper cars. Combined massof car A and its occupants is 250 kg. Combined mass of car Band its occupants is 200 kg. Car A is 15 m away from car Band moving to right at 2.0 m/s when driver decides to bump into carB which is at rest Car A accelerates at 1.5 m/s squared to a speed of 5.0m/s and then continues at constant velocity until it strikes carB Calculate total time for car A to travel the 15 m
After collision car B moves to right at speed of 4.8 m/s Calculate speed of car A after collision What is direction of motion of car A after collision? is this elastic collision? Why/why not? Several students are riding bumper cars. Combined massof car A and its occupants is 250 kg. Combined mass of car Band its occupants is 200 kg. Car A is 15 m away from car Band moving to right at 2.0 m/s when driver decides to bump into carB which is at rest Car A accelerates at 1.5 m/s squared to a speed of 5.0m/s and then continues at constant velocity until it strikes carB Calculate total time for car A to travel the 15 m
After collision car B moves to right at speed of 4.8 m/s Calculate speed of car A after collision What is direction of motion of car A after collision? is this elastic collision? Why/why not?

Explanation / Answer

So here intially car a accelerates to 5m/s and then it it goesat constant velocity.

So for to go to 5m/s

Time taken t=(v-u)/a =(5-2)/1.5 =2sec

Distance covered S=(v*v-u*u)/2a =(25-4)/2*1.5 =21/3=7m

So distance left is 8m

time to amke 8m t =8/5 =1.6sec

Total time =2+1.6 =3.6sec

Using conversion of momentum.Let v be car A final velocity
then

250*5 =200*4.8+250*v

v =(5-4.8*0.8) =1.16m/s
A will also move along right but loses somespeed.

Applying conversation of energy
Intial =250*25/2 =3125

Final =0.5(200*4.8*4.8+250*1.16*1.6) =2536

it's not valid so inelastic collision

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