Question

A body of mass 2.7 kg makes an elastic collision with another body at rest and continues to move in the original direction but with 1/3 of its original speed.

(a) What is the mass of the other body?
(b) What is the speed of the two-body center of mass id the initial speed of the 2.7kg body was 4.0 m/s?

Answer 1

a)

1.35 kg

b)

2.67 ms⁻¹

Step-by-step explanation:

a)

`m_(1)` = mass of first body = 2.7 kg

`m_(2)` = mass of second body = ?

`v_(1i)` = initial velocity of the first body before collision =
`v`

`v_(2i)` = initial velocity of the second body before collision = 0 m/s

`v_(1f)` = final velocity of the first body after collision =

using conservation of momentum equation

`m_(1) v_(1i) + m_(2) v_(2i) = m_(1) v_(1f) + m_(2) v_(2f)\\(2.7) v + m_(2) (0) = (2.7) ((v)/(3) ) + m_(2) v_(2f)\\(2.7) ((2v)/(3) ) = m_(2) v_(2f)\\v_(2f) = (1.8v)/(m_(2))`

Using conservation of kinetic energy

`m_(1) v_(1i)^(2)+ m_(2) v_(2i)^(2) = m_(1) v_(1f)^(2) + m_(2) v_(2f)^(2) \\(2.7) v^(2) + m_(2) (0)^(2) = (2.7) ((v)/(3) )^(2) + m_(2) ((1.8v)/(m_(2)))^(2) \\(2.7) = (0.3) + (3.24)/(m_(2))\\m_(2) = 1.35`

b)

`m_(1)` = mass of first body = 2.7 kg

`m_(2)` = mass of second body = 1.35 kg

`v_(1i)` = initial velocity of the first body before collision = 4 ms⁻¹

`v_(2i)` = initial velocity of the second body before collision = 0 m/s

Speed of the center of mass of two-body system is given as

`v_(cm) = ((m_(1) v_(1i) + m_(2) v_(2i)))/((m_(1) + m_(2)))\\v_(cm) = (((2.7) (4) + (1.35) (0)))/((2.7 + 1.35))\\\\v_(cm) = 2.67` ms⁻¹