Question

23 Two carts on a frictionless track have mass m and 4m respectively. The heavier

cart is travelling 0.78 m/s [E] and less massive cart is travelling 1.3 m/s [W). After a
head-on elastic collision, determine the final velocities of both carts.

Answer 1

• lighter: 2.028 m/s E
• heavier: 0.052 m/s W

Step-by-step explanation:

You want the final velocities of a cart of mass m traveling 1.3 m/s W and a cart of mass 4m traveling 0.78 m/s E after an elastic collision.

Elastic collision

In an elastic collision, both momentum and kinetic energy are conserved. If we let x and y represent the post-collision velocities of the lighter and heavier carts, respectively, then we can write equations ...

m(-1.3) +4m(0.78) = mx +4my . . . . . . . conservation of momentum

m/2(-1.3)² +4m/2(0.78²) = m/2x² +4m/2y² . . . . . conservation of energy

Solution

Dividing the first equation by m, and the second equation by m/2, we can substitute for x:

x = -1.3 +4(0.78) -4y

(-1.3)² +4(0.78)² = (-1.3 +4(0.78) -4y)² +4y²

4.1236 = (1.82 -4y)² +4y²

20y² -14.56y -0.8112 = 0 . . . . simplify to standard form

(y +0.052)(y -0.78) = 0 . . . . . . divide by 20 and factor

y = -0.052 . . . . . . m/s (negative is West)

Then x is ...

x = 1.82 -4(-0.052) = 2.028

The final velocities of the carts are ...

• lighter: 2.028 m/s E
• heavier: 0.052 m/s W