Question

PRACTICE IT . Use the worked example above to help you solve this problem. Two billiard balls of identical mass move toward each other as shown in the figure. Assume that the collision between them is perfectly elastic. If the initial velocities of the balls are v1i = +25.2 cm/s and v2i = −21.8 cm/s, what are the velocities of the balls after the collision? Assume friction and rotation are unimportant. (Indicate the direction with the sign of your answer.) v1f = Correct: Your answer is correct. . cm/s v2f = Correct: Your answer is correct. . cm/s . EXERCISE HINTS: GETTING STARTED | I'M STUCK! . Find the final velocity of the two balls if the ball with velocity v2i = −21.8 cm/s has a mass equal to half that of the ball with initial velocity v1i = +25.2 cm/s. (Indicate the direction with the sign of your answer.) v1f= cm/s v2f = cm/s

Answer 1

ELASTIC COLLISION

m1 = m kg                        m2 = m/2 kg

speeds before collision

v1i = 25.2 cm/s                   v2i = -21.8 cm/s

speeds after collision

v1f = ?                          v2f = ?

initial momentum before collision

Pi = m1*v1i + m2*v2i

after collision final momentum

Pf = m1*v1f + m2*v2f

from momentum conservation

total momentum is conserved

Pf = Pi

m1*v1i + m2*v2i = m1*v1f + m2*v2f .....(1)

from energy conservation

total kinetic energy before collision = total kinetic energy after collision

KEi = 0.5*m1*v1i^2 + 0.5*m2*v2i^2

KEf =   0.5*m1*v1f^2 + 0.5*m2*v2f^2

KEi = KEf

0.5*m1*v1i^2 + 0.5*m2*v2i^2 = 0.5*m1*v1f^2 + 0.5*m2*v2f^2 .....(2)

solving 1&2

we get

v1f = ((m1-m2)*v1i + (2*m2*v2i))/(m1+m2)

v1f = ( ((1-0.5)*25.2) - (2*0.5*21.8) ) /(1 + 0.5) = -6.13 cm/s

v2f = ((m2-m1)*v2i + (2*m1*v1i))/(m1+m2)

v2f = ( -((0.5-1)*21.8) + (2*1*25.2) ) /(1+0.5)

v2f = 40.8 cm/s