Question

7. A 3 000 kg truck runs into the rear of a 1 000 kg car that was stationary. The truck and car are locker together after the collision and move with speed 9 m/s. What was the speed of the truck before the collision? 

8. In the previous question determine if the collision was an elastic collision.

Answer 1

According to momentum conservation,

momentum of truck= momentum of ( car+ truck)

Let v is the initial velocity of truck.

3000* v = (1000+3000) *9

v= 4000*9/3000

V= 12 m/ s

Is the velocity of truck earlier.

8.

In elastic collision energy is conserved

So 1/2Mv²= 1/2(M+m)*(9*9)

0r,

v²=( 4000/3000)*9*9

v=10.39 m/s in case of elastic collision.

Answer 2

SOLUTION :

7.

Let speed of the truck be V m/s before collision.

Momentum before collision = Momentum after collision

=> 3000 * V + 1000 * 0 = (3000+1000) * 9

=> V = (4000 * 9) / 3000

=> V = 12 m/s 

S0, the speed of the truck before collision is 12 m/s m/s (ANSWER)

8.

In elastic collision, K. E. Remains unchanged.

Let speed of the truck be V m/s before elastic collision.

=> K.E. before collision = K.E. after collision

=> 1/2 * 3000 * (V)^2 + 1/2 * 1000 * 0^2 = 1/2 (3000 + 1000) * 9^2

=>  1500 V^2 = 162000

=> V = sqrt(162000 / 1500)

=> V = 10.39 m/s. 

S0, the speed of the truck before elastic collision is 10.39 m/s m/s (ANSWER)