Question

A 2.7-kg block is traveling in the −x direction at 5.6 m/s, and a 1.2-kg block is traveling in the +x direction at 3.1 m/s.

(a) Find the velocity vcm of the center of mass. _____m/s

(b) Subtract vcm from the velocity of each block to find the velocity of each block in the center-of-mass reference frame. 2.7-kg block _____m/s and 1.2-kg block _____m/s

(c) After they make a head-on elastic collision, the velocity of each block is reversed (in the center-of-mass frame). Find the velocity of each block in the center-of-mass frame after the collision. 2.7-kg block _____m/s and 1.2-kg block _____m/s

(d) Transform back into the original frame by adding vcm to the velocity of each block. 2.7-kg block _____m/s and 1.2-kg block _____m/s

Answer 1

Given

Mass ofthe blocks

m1= 2.7 kg travelling at, v1=5.6 m/s(-i)

m2=1.2 kg travelling at, v2=3.1 m/s(i)

(a) The velocity v_cm of the center of mass is

v_cm =  

=

v_cm = 4.83 m/s ()

(b) The velocity of each block in the center of mass reference is

For m₁.v₁'= v1-v_cm = (-5.6m/s)- (-4.83m/s)

v₁' = (-0.77m/s)

For m2 . v₂' = v₂  - v_cm = (3.1m/s) - (-4.83m/s)

v₂' = (7.93m/s)

(c) The velocity of the block reversed after the collision.

now the velocity of each block is

v₁'' = (0.77m/s)

v₂'' = (7.93m/s)

(d) Transforming back into the original frame,

the velocity of each block is

u₁' = v₁'' + v_cm = (0.77m/s)+(-4.83m/s)

u₁' = (-4.06m/s)

u₂' = v₂'' + v_cm = (-7.93m/s)+ (-4.83m/s)

u₂' = (-12.73m/s)