Answered

A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s.

After the collision, the 2.00-kg object has a velocity 5.00 m/s to the west. How much kinetic energy

was lost during the collision?

Verified Answer :

meerkat18

Momentum = mass x velocity

Before collision
Momentum 1 = 2 kg x 20 m /s = 40 kg x m/s
Momentum 2 = 3 kg x -10m/s = -30 kg x m/s

After collision
Momentum 1 = 2 kg x -5 m/s = -10 m/s
Momentum 2 = 3 kg x V2 = 3V2

Total momentum before = total momentum after
40 + -30 = -10 + 3V2
V2 = 6.67 m/s

Total kinetic energy before
= (1/2) [ 2 kg 20 m/s  2 + 3 kg  ( -10 m/s) 2 ]
= 550 J

Total kinetic energy after
= (1/2) [ 2 kg ( - 5 m/s)  2 + 3 kg  6.67 m/s 2 ]
= 91.73 J

Total kinetic energy lost during collision
=550 J - 91.73 J
= 458.27 J

littleredstudies

Kinetic energy before:

E = 0.5mv^2 + 0.5mv^2

= 0.5220^2 + 0.53(-10)^2

= 400 + 150

= 550J

Kinetic energy after:

E = 0.5mv^2

= 0.52(-5)^2

= 25J

Kinetic energy lost during the collision:

E = 550 - 25

= 525J

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