A 50.0kg object is moving east at an unknown velocity when it collides with a 60.0kg stationary object. After the collision, the 50.0kg is traveling at a velocity of 6.0m/s 50.0degree Nof E and the 60.0kg object is traveling at a velocity of 6.3m/s 38 degree Sof E. What was the velocity of the 50.0kg object before the collision?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

This problem can be solved by using the definition of the linear momentum: `vecp = m*vecV` (mass times velocity) and the law that momentum of a system in the absense of external forces is conserved.

According to the conservation of linear momentum law, the vector linear momentum of the 50...

Unlock
This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial

This problem can be solved by using the definition of the linear momentum: `vecp = m*vecV` (mass times velocity) and the law that momentum of a system in the absense of external forces is conserved.

According to the conservation of linear momentum law, the vector linear momentum of the 50 kg object before the collision is the same as the total vector linear momentum of 50 kg and 60 kg objects after the collision. Consider the components of the vector momentum along the east-west line before and after the collision:

Before the collision the 60 kg object is stationary, it has no linear momentum. The total momentum is

`p_1 = 50kg*V` , where V is the velocity of 50 kg object before the collision.

After the collision the component of total momentum of both objects along east-west line is

`p_2 = 50kg*6 m/s *cos50 + 60kg*6.3 m/s*cos38 = 490.7 kg*m/s`

Since `p_1 = p_2` , `50kg*V = 490.7 kg*m/s` and

`V=9.8 m/s`

The velocity of 50 kg object before the collision is 9.8 m/s.

Approved by eNotes Editorial Team