Question

collision is perfectly elastic, what is the change in the kinetic energy of the 620-g object? A 620-g object traveling at 2.1 m/s collides head-on with a 320-g object traveling in the opposite direction at 3.8 m/s. If the collision is perfectly elastic, what is the change in the kinetic energy of the 620-g object? It loses 0.23 J. It loses 1.4 J. It gains 0.69 J. It loses 0.47 J. It doesn't lose any kinetic energy because the collision is elastic.

Answer 1

Answer:

The answer is: It loses 0.23 J

Explanation:

When the collision is elastic, both, momentum and kinetic energy is conserved, thus, the velocity is equal:

$v_{1} =(\frac{m_{1}-m_{2}}{m_{1}+m_{2} } )u_{1}+\frac{2m_{2}u_{2}}{m_{1}+m_{2}}$

Where

m₁ = 620 g = 0.62 kg

m₂ = 320 g = 0.32 kg

u₁ = 2.1 m/s

u₂ = -3.8 m/s

Replacing:

$v_{1} =(\frac{0.62-0.32}{0.62+0.32} )*2.1+\frac{2*0.32*(-3.8)}{0.62+0.32} =-1.917m/s$

The change of kinetic energy is:

$E_{k} =\frac{1}{2} m*delta-v^{2} =\frac{1}{2} *0.62*((-1.917)^{2}-(2.1)^{2} )=-0.228=-0.23J$

The negative sign indicates a loss of energy

Answer 2

Answer:

It doesn't lose any kinetic energy because the collision is elastic.

Explanation:

In an elastic collision, the momentum and kinetic energy are conserved