You throw a 0.52-kg target upward at 15 m/s. When it is at a heigh of 10 m above

Question

You throw a 0.52-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 25 m/s upward. Assume the interaction is instantaneous.

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

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

*I tried adding up the kinetic energies plugging that value into the kinetic equation solving for the v and my final answer was 19.44 but it said the answer was wrong? could someone tell me what I did wrong?*

Explanation / Answer

A) let the speed of target is u m/s downwards

using the kinemtic equation

u^2 - 15^2 = -2*10 * 9.8

u = 5.4 m/s

here During this inelastic collision, if we assume V as final speed

v*(0.338 + 0.52) = 0.338 * 25 - 5.4 * 0.52

v = 6.57 m/s upwards ----------------<<<<<<<<<<<<<<<Answer

--------------------------------------------------------------

B)

let the speed of combnation before hitting ground is vf

using third equation of motion ,

Vf^2 - U^2 = 2*a*s

vf^2 - 6.57^2 = 2 * 9.8 * 10

vf = 15.46 m/s----------------------<<<<<<<<<<<<<Answer

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