In the movie for this pre-session, we analyzed a one-dimensonal elastic collisio

Question

In the movie for this pre-session, we analyzed a one-dimensonal elastic collision between a basketball (mass 3m) and a baseball (mass m). The balls had equal-and-opposite velocities before the collision, and after the collision the basketball was at rest and the baseball had a velocity twice the magnitude it had initially, but in the opposite direction.

Let's analyze a similar one-dimensonal elastic collision between a red ball (mass 7m) and a blue ball (mass m). Again, the balls have equal-and-opposite velocities before the collision - let's say each ball has a speed of 2.00 m/s before the collision.

(a) Calculate the speed of the red ball after the collision.

(b) Calculate the speed of the blue ball after the collision.

Explanation / Answer

In collision the momentum and kinetic energy (KE) is conserved So,

Taking one direction is positive so that the direction of velocity of other ball is negative

Initial momentum = Final Momentum

Initial KE = Final KE

m1u1 + m2(-u2) = m1v1 + m2v2

m1u1 - m2u2 = m1v1 + m2v2

here, u1 = u2 = 2 m/s

(7m x 2) - (2 m) = 7m v1 + m v2

12 m = m (7v1 + v2)

12 = 7v1 + v2 ---------------------------equation 1

KE Conservation
1/2 m1u1^2 - 1/2 m2u2^2 = 1/2 m1v1^2 + 1/2 m2v2^2

(7m x 4) - (4 m) = 7m v1^2 + m v2^2

24 m = m (7 v1^2 + v2^2)

24 = 7 v1^2 + v2^2 ---------------------equation 2

From eq. 1
v2 = 12 - 7 v1 (Put it in eq.2)

24 = 7 v1^2 + (12 - 7v1)^2

Square root both the sides

4.89 = 2.64 v1 + (12 - 7v1)

4.89 = -4.36 v1 + 12

{v1 = 1.63 m/s} Answer a

Put this in eq 1

12 = 7v1 + v2

12 = (7 x 1.63) + v2

12 = 11.41 + v2

{v2 = 0.59 m/s} Answer b

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