MIDTERM 2 SECTION NAME- 3. The boy shoots a smaller ball (ball 1) with mass of m
ID: 1783787 • Letter: M
Question
MIDTERM 2 SECTION NAME- 3. The boy shoots a smaller ball (ball 1) with mass of m/s straight to the first ball (ball 2) with mass of 0.200 kg at rest on the Rooftop simplicity lets assume one-dimensional motion. Ball 1 is moving east toward the 2. Two balls elastically collide and they bounce off each other. 0.100 kg at initial speed of 3 is flat and the ball was launched straight towa rd the other ball, so for (a) Find the total momentum of two balls before collision. (4 points) Answer: kg m/s (b) Find the total kinetic energy of two balls before collision. (4 points) e) Using conservation of momentum and kinetic energy, find each ball's velocity after collision. (4 points) Hint: Ball 1 or Ball 2's velocity after collision is not 0 m/s) Answer: BallL -m/s Direction Ball 2 m/s DirectionExplanation / Answer
a)
total moemtum beofore collision Pi = m1*v1i + m2*v2i
Pi = (0.1*3) + (0.2*0) = 0.3 kg m/s
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(b)
kinetic energy before sollision Ki = (1/2)*m1*v1i^2 + (1/2)*m2*v2i^2
Ki = (1/2)*0.1*3^2 + 0 = 0.45 J
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momentum after collision Pf = m1*v1f + m2*v2f
kinetic energy after collision Kf = (1/2)*m1*v1f^2 + (1/2)*m2*v2f^2
from momentum conservation
Pf = Pi
0.1*v1f + 0.2*v2f = 0.3
v1f + 2*v2f = 3
v2f = (3-v1f)/2
from conservation of kinetic energy Kf = Ki
(1/2)*0.1*v1f^2 + (1/2)*0.2*v2f^2 = 0.45
0.1*v1f^2 + 0.2*v2f^2 = 0.9
v1f^2 + 2*v2f^2 = 9
v1f^2 + (3-v1f)^2/2 = 9
velocity of ball 1 v1f = -1 m/s <<------------ANSWER
velocity of ball 2 v2f = (3-(-1))/2 = 4/2 = 2 m/s <<<------------ANSWER
ball1 1 m/s direction -xaxis (west)
ball2 2 m/s direction xaxis (east)
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