An automobile engine propels a 1000 kg car (A) along a levelled road at a spedd of 36 km//h. Find the power if the opposing frictional force is 100 Now, suppose after travelling a distance of 200 m, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stop at the same time. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of the car (B) just after the collision.

Explanation

Given, mass of car (A) = 1000kg
Mass of car B = 1000kg
Force applied by car A = 100 N
`(v_(A))"Speed of car"(A)=36kmh^(-1)=36xx(5)/(18)=10ms^(-1)[:'1km//h=(5)/(18)m//s]`
`:."Power of car"(A),P_(A)=F.v_(A)=100xx10=1000W`
`["F=force exerted by the car underset(-)A against friction"]`
Again, for car (A) Newton's low, F = ma
`100=1000xxarArra=(100)/(1000)`
`a=(1)/(10)ms^(-1)`
Velocity of car A after travelling 200 m is given by
From third equatioin of motion, `v^(2)=u^(2)+2as" "["for c ar A here", u=10ms^(-1),s=200m]`
`v^(2)=(10)^(2)+2xx(1)/(10)xx200`
`v^(2)=100+40=140`
`v=sqrt140=11.8ms^(-1)`
According to the question, after moving of 200m, the speed of car A, `u_(1)=11.8ms^(-1)`
Just after the collision, the final speed of car A, `v_(1)=0`
before collision, the initial speed of car B, `u_(2)=0`
From conservation of linear momentum,
`m_(1)u_(1)+m_(2)u_(2)=m_(1)v_(1)+m_(2)v_(2)`
[Let just after the after the collision, the speed of the car B is `v_(2)`]
`m_(1)xx11.8m_(1)+m_(2)xx0=m_(1)xx0+m_(2)xxv_(2)`
`11.8m_(1)=m_(2)v_(2)" "[As:'m_(1)=m_(2)]`
`11.8m_(1)=m _(1)v_(2)`
`rArrv_(2)=11.8ms^(-1)`