Suppose a 1.5 kg watermelon is hung from a string. A bolt with a mass of 30 gram

Question

Suppose a 1.5 kg watermelon is hung from a string. A bolt with a mass of 30 grams is shot into the watermelon. Before impact the bolt has a velocity of 45.0 m/s. After exiting the watermelon its velocity is 15.0 m/s. The string is hanging straight down, therefore there is no horizontal force on the system when the arrow enters and exits.

Find the speed of the watermelon immediately after the arrow exits.
Determine how much kinetic energy is lost in the collision.
As the watermelon swings on the string, it will rise in height. Find how far in height it rises before stopping.

Explanation / Answer

conservation of momentum : 30*45 = 30*15 + (1500)V V = 900/1500 = 3/5 m/sec change in K.E = (1/2)0.030 * 45*45 - (1/2)(0.03)(225) - (1/2)(1.5)(9/25) : (1/2) [ 1800 ] (0.03) - (1/2)[ 27/50 ] : 27 - 0.27 = 26.73 Joule ~~~ change in KE = chnage in PE (1/2)v*v = gh h = 9/50*g = 1.83 cm answer !!!

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