Question

You throw a 0.42-kg target upward at 15 m/s. When it is at a heigh of 10 m above the launch position and moving downward, it is struck by a 0.338-kg arrow going 27 m/s upward. Assume the interaction is instantaneous.

Part A

What is the velocity of the target and arrow immediately after the collision?

Part B

What is the speed of the combination right before it strikes the ground?

Answer 1

max height obtain by target = u^2 / 2*g = 15² / 2*9.8 = 11.4795

so left height it travel ( h) = 11.4795 - 10 = 1.4795 so velocity above 10 m from launch

apply v^2 = u^2 + 2*g*h

v =

now apply momentum conservation

m1v1 +m2v2 = (m1+m2)V

0.42(-5.4131) + 0.338*(+27) = (0.42+0.338)V

Part A

V = 9.040 up ward

Part B

now total mass wiil go up with this velocity so max hight after that collision point

H_mx  = u²  / 2*g = 9.040²  / 2*9.8 = 4.1694 m

total height from ground = 10+4.1694 = 14.1694 m

now apply

V²  =u² + 2*g*s to calculate velocity before combined mass hit the ground

V²   = 0 +2*9.8 *14.1694

so

V= 16.664 m/s