Question

You throw a penny (m = 2.5 g) with an initial speed of 9.4 m/s at 25◦ above the horizontal. At the maximum height of your thrown penny’s path another penny, traveling straight upward, collides and bounces off of your penny. The speed of the second penny at the moment of collision was 4.1 m/s. a) Determine the maximum height your thrown penny reaches as a result of this collision. b) How far, horizontally, does your penny travel before reaching the ground (assume that the penny was thrown from the ground). Hint: Make sure to account for BOTH directions at ALL times.)

Answer 1

a) at maximum heigt vertical velocity of penny is zero . it only have horizontal velocity that is 9.4cos25

velocity of thrown penny at max. height = 9.45cos25 i = 8.52 m/s i

velocity of other penny = 4.1 m/s j

velocity of other penny after collision = -4.1 j m/s

after collision thrown will also got 4.1 + 4.1 = 8.2 m/s in upward direction.

in vertical using v^2 - u^2 = 2ad

0 - 8.2^2 = 2 x -9.81 x h1

h1 = 3.43 m

height at collision, h2 = (9.4sin25)^2 / 2x9.81 =0.813 m

So max height = h1 + h2 = 4.24 m

b) horizontal distance of collision from throwing position = Range /2 = v^2 sin(2*25) / 2x9.81

= 9.4^2 sin50 / 2x9.81 = 3.45 m

after that penny have to go 0.813m vertical downwards with initial vertical velocity 8.2 m/s

using h = ut + at^2 /2

-0.813 = 8.2t - 9.81t^2 /2

4.905t^2 - 8.2t - 0.813= 0

t = 1.76 sec

horizontal distance traveled in this time = (9.4cos25) x1.76 =15m

total distance = 15 + 3.45 =18.45 m