A penny (m=2.5 g) is thrown with initial speed 9.4 m/s at 25 degrees above the h

Question

A penny (m=2.5 g) is thrown with initial speed 9.4 m/s at 25 degrees above the horizontal. At the maximum height of the penny thrown, another penny traveling straight upwards with a speed of 4.1 m/s collides and bounces off of the thrown penny. a) Determine the max height of the thrown penny as a result of the collision b) How far horizontally does the thrown penny travel before reaching the ground?

(Assuming the penny is of equal mass to the first and the collision is elastic as the second penny bounces off the first penny. )

Explanation / Answer

initial vertical component of thrown penny = 9.4sin25 = 3.973 m/s

initial horizontal component of thrown penny = 9.4cos25 = 8.519 m/s

With this vertical speed height reached before collision is V_vertical2/(2*g) = 3.973*3.973/(2*9.8) = 0.805 m

At this height thrown penny got 4.1 m/s upwardes because of collision

So height reached by this speed = 4.1*4.1/(2*9.8) = 0.858 m

So max height = sum of both height =  0.858 + 0.805 m = 1.6626 m Answer (A)

B) Distance moved = time*horizontal velocity

Applying V = U + a*t

So t = *(9.4+4.1)/9.8

Time = 2*t = 2.755 sec

So horizontal distance = 2.755*8.519 = 23.5 m Answer

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