Question

Two carts undergo an inelastic collision where they stick together. Cart A has an initial velocity v0, and the second cart B is initially at rest. After the collision, it is observed that the ratio of the final kinetic energy system to its initial kinetic energy is KfK0= 1/6. Determine the ratio of the carts' masses, mBmA. (Assume the track is frictionless.)

Answer 1

Utilizing the conservation of momentum and the given kinetic energy ratio, we can derive the final velocity in terms of the initial velocity and solve for the mass ratio mB/mA after an inelastic collision on a frictionless surface.

Step-by-step explanation:

To determine the ratio of the carts' masses mB/mA when two carts undergo an inelastic collision, we can utilize the principle of conservation of momentum along with the given ratio of final kinetic energy to initial kinetic energy, Kf/K0 = 1/6.

The initial momentum of the system is p0 = mAv0 because cart B is at rest. After the collision, the combined mass (mA + mB) has a final velocity vf, so the final momentum is pf = (mA + mB)vf. By conservation of momentum, p0 = pf.

The initial kinetic energy is K0 = 1/2 mAv02, and the final kinetic energy is Kf = 1/2 (mA + mB)vf2. Given that Kf/K0 = 1/6, we can find vf in terms of v0 and then solve for the mass ratio mB/mA.

Let's solve for the final velocity first: Kf = K0/6 => 1/2 (mA + mB)vf2 = 1/12 mAv02. From conservation of momentum, we have mAv0 = (mA + mB)vf, which allows us to eliminate vf and solve for mB/mA.

Answer 2

Step-by-step explanation:

Initial kinetic energy of the system = 1/2 mA v0²

If Vf be the final velocity of both the carts

applying conservation of momentum

final velocity

Vf = mAvo / ( mA +mB)

kinetic energy ( final ) = 1/2 (mA +mB)mA²vo² / ( mA +mB)²

= mA²vo² / 2( mA +mB)

Given 1/2 mA v0² / mA²vo² / 2( mA +mB) = 6

mA v0² x ( mA +mB) / mA²vo² = 6

( mA +mB) / mA = 6

mA + mB = 6 mA

5 mA = mB

mB / mA = 5 .