You and your friends are doing physics experiments on a frozen pond that serves
ID: 2206665 • Letter: Y
Question
You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 80.0 , is given a push and slides eastward. Abigail, with mass 56.0 , is sent sliding northward. They collide, and after the collision Sam is moving at 38.0 north of east with a speed of 6.20 and Abigail is moving at 21.0 south of east with a speed of 8.20 . 1.What was the speed of each person before the collision? Sam's speed: Abigail's speed: 2.By how much did the total kinetic energy of the two people decrease during the collision?Explanation / Answer
mS = 80 kg; mA = 56 kg; S1 = 0 degrees; A1 = 90 degrees; vS2 =6.2 m/2; vA2 = 8.2 m/s; S2 = 38 degrees; A2 = -21 degrees
(1) Conservation of linear momentum: The total (sum of) momentum before the collision (denoted by subscripts of 1) equals the total momentum after the collision (denoted by 2).
In the x direction:
mSvS1cosS1+mAvA1cosA1=mSvS2cosS2+mAvA2cosA2
80 vS1cos(0) + 56 vA1cos(90) = (80)(6.2)cos(38) + (56)(8.2)cos(-21) [cos 0 = 1; cos 90 = 0]
80 vS1 = 390.85 + 428.7 = 819.55
vS1 = 819.55/80 = 10.24 m/s
In the y direction:
mSvS1sinS1+mAvA1sinA1=mSvS2sinS2+mAvA2sinA2
80 vS1sin(0) + 56 vA1sin(90) = (80)(6.2)sin(38) + (56)(8.2)sin(-21) [sin 0 = 0; sin 90 = 1]
56 vA1 = 305.37 - 164.56 = 140.81
vA1 = 140.81/56 = 2.51 m/s
(2)
KE1-KE2
[(1/2)mSvS12 + (1/2)mAvA12] - [(1/2)mSvS22 + (1/2)mAvA22]
Simplifying:
mS(vS12 - vS22 ) + mA(vA12 - vS22 )
You should be able to plug in the numbers to get the value.
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