Asked by Lana
I've been stuck on this questions for 2 days
A 7.2-kg bowling ball traveling at a speed of 3.5 m/s collides with a 1.2-kg bowling pin at rest. After the collision, the pin moves with a speed of 3.0 m/s at an angle of 60° with respect to the original direction of motion of the bowling ball. What is the final velocity of the ball (magnitude and direction)? Is the collision approximately elastic?
A 7.2-kg bowling ball traveling at a speed of 3.5 m/s collides with a 1.2-kg bowling pin at rest. After the collision, the pin moves with a speed of 3.0 m/s at an angle of 60° with respect to the original direction of motion of the bowling ball. What is the final velocity of the ball (magnitude and direction)? Is the collision approximately elastic?
Answers
Answered by
Scott
momentum is conserved in both the x and y directions
the initial momentum is all in the x direction -- the moving ball
PMy = 3.0 m/s * sin(60º) * 1.2 kg
BVy = PMy / 7.2
PMx = 3.0 m/s * cos(60º) * 1.2 kg
BVx = 3.5 - (PMx / 7.2)
use Pythagoras to find BV
the initial momentum is all in the x direction -- the moving ball
PMy = 3.0 m/s * sin(60º) * 1.2 kg
BVy = PMy / 7.2
PMx = 3.0 m/s * cos(60º) * 1.2 kg
BVx = 3.5 - (PMx / 7.2)
use Pythagoras to find BV
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