Question

An object with mass mAmA = 1.6 kgkg , moving with velocity v⃗ A=(4.2iˆ+5.6jˆ−3.0kˆ)m/sv→A=(4.2i^+5.6j^−3.0k^)m/s, collides with another object of mass mBmB = 4.2 kgkg , which is initially at rest. Immediately after the collision, the object with mass mAmA = 1.6 kgkg is observed traveling at velocity v⃗ ′A=(−2.0iˆ+3.0kˆ)m/sv→′A=(−2.0i^+3.0k^)m/s.   Find the velocity of the object with mass mBmB after the collision. Assume no outside force acts on the two masses during the collision.

Enter the xx, yy, and zz components of the velocity separated by commas. Express your answer to two significant figures.

Answer 1

Given before collision

mA = 1.6 kg ,UA = (4.2iˆ+5.6jˆ−3.0kˆ) m/s   

mB = 4.2 kg , UB = 0 because second body is at rest initially

after collision the body of mass mA  = 1.6 kg is moving with the velocity VA  = (−2.0iˆ+3.0kˆ)m/s

Let VB be final velocity of the second body

Applying conservation of momentum,

mA UA + mBUB = mA VA + mBVB

1.6 (4.2iˆ+5.6jˆ−3.0kˆ) + 0 = 1.6  (−2.0iˆ+3.0kˆ) + 4.2 VB

1.6 (4.2iˆ+5.6jˆ−3.0kˆ +2.0iˆ - 3.0kˆ) = 4.2 VB

1.6 (6.2iˆ+5.6jˆ−6.0kˆ) = 4.2 VB

8 (6.2iˆ+5.6jˆ−6.0kˆ) = 21 VB

VB = 2.36 iˆ + 2.13 jˆ − 2.28 kˆ

The respective velocity components of B alon X, Y and Z axes after collision are 2.36, 2.13, - 2.28