# Answer on Mechanics | Relativity Question for prack

Question #50084

This question refers to: Laws of motion

A mass of 2 kg travelling at 2 m/s is struck by a mass of 4 kg travelling at 3 m/s. If the coefficient of restitution for the two materials is 0.4, calculate their final velocities.

Give the final velocity value in m/s of the 4kg mass here:

A mass of 2 kg travelling at 2 m/s is struck by a mass of 4 kg travelling at 3 m/s. If the coefficient of restitution for the two materials is 0.4, calculate their final velocities.

Give the final velocity value in m/s of the 4kg mass here:

Expert's answer

Given:

m1=2

v1b=2 (velosity before collision)

m2=4

v2b=3 (velosity before collision)

cr=0.4 (coeffitient of restitution)

We want to find v1a (velosity after collision)-?

Solution:

According ofComversation of the impulse

m1*v1b - m2*v2b = - m1*v1a + m2*v2a

and

definition of coeffitient of restitution

cr= (v1b-v1a)/(v2a-v2b)

we have

v1a= [(m1*v1b-m2*v2b)-(m2*v2+m2*v1b/cr)]/[m1-m2/cr]=4 m/s

v2a=v2b+(v1b-v2a)/cr=-2 m/s

Answer: v1a=3.4 m/s v2a=-2 m/s

m1=2

v1b=2 (velosity before collision)

m2=4

v2b=3 (velosity before collision)

cr=0.4 (coeffitient of restitution)

We want to find v1a (velosity after collision)-?

Solution:

According ofComversation of the impulse

m1*v1b - m2*v2b = - m1*v1a + m2*v2a

and

definition of coeffitient of restitution

cr= (v1b-v1a)/(v2a-v2b)

we have

v1a= [(m1*v1b-m2*v2b)-(m2*v2+m2*v1b/cr)]/[m1-m2/cr]=4 m/s

v2a=v2b+(v1b-v2a)/cr=-2 m/s

Answer: v1a=3.4 m/s v2a=-2 m/s

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