A ball with a mass of 0.625 is initially at rest. It is struck by a second ball

Question

A ball with a mass of 0.625 is initially at rest. It is struck by a second ball having a mass of 0.425 , initially moving with a velocity of 0.240 toward the right along the axis. After the collision, the 0.425 ball has a velocity of 0.170 at an angle of 36.6 above the axis in the first quadrant. Both balls move on a frictionless, horizontal surface.
a.)What is the magnitude of the velocity of the 0.625 ball after the collision?
b.)What is the direction of the velocity of the 0.625 ball after the collision?
c.)What is the change in the total kinetic energy of the two balls as a result of the collision?

Explanation / Answer

a) conservation of momentum along X direction

the horizontal components of velocity after the collision will be v1cos1 and v2cos2

m1u1+m2u2= m1v1cos1+m2v2cos2

0.625*0+0.425*0.24= 0.625*v1cos1 + 0.425*0.17*cos36.6

0.102=0.625*v1cos1 + 0.058

v1cos1=0.102-0.058/0.625= 0.0704 m/sec

conservation of momentum along Y direction

vertical components will be v1sin1 and v2 sin2

m1*v1sin1= m2*v2sin2

v1sin1= 0.425*0.17*sin36.6/0.625= 0.0689

v1=sqrt[(v1cos1)^2+(v1sin1)^2]

v1= sqrt(0.0097)= 0.0985 m/sec

b) v1sin1= 0.0689

v1cos1= 0.0704

taking ratio

tan1= .9787

1= arctan(.9789)= 44.38 degrees

the angle is 44.38 degrees below the axis

c) KE1= KE1i-KE1f= 0.5*0.625*(0.0985)^2= 3.031*10^-3 J

KE2= KE2i-KE2f= 0.5*0.425(0.24^2-0.17^2)= 6.099*10^-3 J

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