You and your friends are doing physics experiments on a frozen pond that serves

Question

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 70.0kg , is given a push and slides eastward. Abigail, with mass 59.0kg , is sent sliding northward. They collide, and after the collision Sam is moving at 36.0? north of east with a speed of 7.00m/s and Abigail is moving at 21.0? south of east with a speed of 8.60m/s .

A) What was the speed of each person before the collision?
Sam's speed?

B) Abigail's speed?

C)By how much did the total kinetic energy of the two people decrease during the collision?

Explanation / Answer

Ms = mass of sam = 70.0 kg
Vsi = initial velocity sam = east = x positive

Ma = mass of abigail = 59.0 kg
Vai = initial velocity of abigail = north = y positive

when they meet they dont stick so this is elastic collision

mass's dont change

Vsf = final velocity sam = 7.0 m/s 36 degs from the x axis

Vaf = final velocity abigail = 8.60 m/s -21 degs from the x axis

conservation of momentum is different for x and y because they are perpendicular

so find the components

Vsfx = 7.0* cos 36
Vsfy = 7.0 * sin 36

Vafx = 8.60* cos 21
Vafy = 8.60*sin 21

now we can do momentum of x and momentum of y separate

x direction only sam is initially moving x

Vsi * Ms = Ms*Vsfx + Ma*Vafx

now solve for Vsi

Vsi = (Ms*Vsfx + Ma* Vafx)/ Ms

Vsi = Vsfx + Ma*Vafx/Ms

putting in values

Vsi = 7.0*cos36 + (59.0/70.0)*8.60*cos 21
Vsi = 12.43 m/s east

now for y motion only abigail is moving y direction initially

Vai *Ma= (Ms*Vsfy + Ma*Vafy)

solve for Vai

Vai = (Ms*Vsfy + Ma*Vafy)/Ma
Vai = Vafy + (Ms/Ma)*Vsfy

Vai = 8.60*sin 21 +(70.0/59.0)*7.0*sin 36
Vai = 7.963 m/s north

so does that help you solve these types of problems

we know Ke now for both before and after

Kesi = 0.5* 70.0* 12.43^2 = 5407.6715 J

Kesf = 0.5*70.0*7.0^2 = 1715 J
delta Kes = 1715 - 5407.6715 = -3692.6715 J
it decreased for Sam

Keai = 0.5* 59.0* (7.963)^2
Keai = 1870.57 J

Keaf = 0.5*59.0*8.6^2
Keaf = 2181.82 J

delta Kea = 2182.82 - 1870.57 = 311.25 J
so it increased for Abigail

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