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1 month ago

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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? 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.

2 answers:

Sav [3.1K]1 month ago

7 0

Answer:

The result is: It decreases by 0.23 J

Explanation:

In an elastic collision, both momentum and kinetic energy are preserved, leading to equal velocities:

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

Where

m₁ = 620 g = 0.62 kg

m₂ = 320 g = 0.32 kg

u₁ = 2.1 m/s

u₂ = -3.8 m/s

Substituting:

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

The change in kinetic energy is:

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

A negative value signifies energy loss

Keith_Richards [3.2K]1 month ago

6 0

Answer:

No kinetic energy is lost as the collision is elastic.

Explanation:

Throughout an elastic collision, both momentum and kinetic energy remain conserved.

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