Question

A 12-kg bowling ball is rolled toward a stationary, 0.9-kg bowling pin at 2.4m/s to the right. After the collision, the bowling ball continues to move in the same direction at 1.3m/s.

Determine the final velocity of the bowling pin.

Answer 1

14.6m/s

Step-by-step explanation:

elastic collision formula

mv1+mv2=mv1'+mv2'

mv1= mass times velocity of the first object

mv2= mass times velocity of the second object

mv1'=mass times velocity of the first object after the collision

mv2= mass times velocity of the second object after the collision.

we know that the pin is not moving so we multiply 0 or its velocity by the mass 0.9kg this equals zero so we don't add this part to the equation.

mv1=mv1'+mv2'

plugin the values and solve using order of operations

12kg(2.4m/s)=12kg(1.3m/s)+0.9kg(v2')

28.8kg.m/s=15.6kg.m/s+0.9kg(v2')

at this point you need to move 15.6kg.m/s to the other side of the equation so subtract 15.6kg.m/s to 28.8kg.m/s

28.8kg.m/s=15.6kg.m/s+0.9kg(v2')

28.8kg.m/s - 15.6kg.m/s= 0.9kg(v2')

finally Divide by 0.9kg in order to leave v2' in one side.

13.2kg.m/s=0.9kg(v2')

13.2kg.m/s÷0.9kg=v2'

14.6kg.m/s=v2'