A truck moving with an initial speed v collides with a car that is one-half as massive as the incident truck and is initially at rest. After the collision, the truck moves off in the same direction as before the collision with a speed v/3. Determine the following:
(a) the final speed of the car in terms of the initial speed of the truck
(b) if the collision is elastic or inelastic

a. 4v/3  b. The collision is elastic.

Step-by-step explanation:

(a) the final speed of the car in terms of the initial speed of the truck

From the law of conservation of momentum, the initial and final momentum of the system is the same.

So, Mv + mV = Mv' + mV' where M = mass of truck, v = initial velocity of truck, v' = final velocity of truck = v/3, m = mass of car = M/2, V = initial velocity of car = 0 (since it was initially at rest) and V' = final velocity of car.

So, substituting the values of the variables into the equation, we have

Mv + mV = Mv' + mV'

Mv + M/2 × 0 = Mv/3 + (M/2)V'

Mv + 0 = Mv/3 + (M/2)V'

Mv - Mv/3 = (M/2)V'

2Mv/3 = (M/2)v'

v' = 2Mv/3 ÷ M/2

v' = 4v/3

(b) if the collision is elastic or inelastic

The collision is elastic if the initial kinetic energy equals the final kinetic energy.

So, initial kinetic energy K = initial kinetic energy of truck + initial kinetic energy of car = 1/2Mv² + 0 = 1/2Mv²

Final kinetic energy K' = final kinetic energy of truck + final kinetic energy of car = 1/2Mv'² + 1/2mV'²

= 1/2M(v/3)² + 1/2(M/2)(4v/3)²

= Mv²/(2 × 3²)(1 + 4²/2)

= (Mv²/18)(1 + 16/2)

= (Mv²/18)(1 + 8)

= 9Mv²/18

= Mv²/2

Since K = 1/2Mv² = K' = 1/2Mv², the collision is elastic.

true? is that what you're asking or do you mean something else?

900 degrees

step-by-step explanation: