Question

1. A toy car with mass m1 travels to the right on a frictionless track with a speed of 3 m/s. A second toy car with mass m2 travels with a speed of 6 m/s to the left on the same track. The two cars collide and stick together. What are the cars’ speeds after the collision?

Answer 1

Answer:

v = 3(m1 - 2m2)/(m1 + m2)

Explanation:

Parameters given:

Velocity of first toy car with mass m1, u1 = 3 m/s (taking the right direction as the positive axis)

Velocity of second toy car with mass m2, u2 = -6 m/s (taking the left direction as the negative x axis)

Using conservation of momentum principle:

Total initial momentum = Total final momentum

m1*u1 + m2*u2 = m1*v1 + m2*v2

Since they stick together after collision, they have the same final velocity.

m1*3 + (m2 * -6) = m1*v + m2*v

3m1 - 6m2 = (m1 + m2)v

v = (3m1 - 6m2) / (m1 + m2)

v = 3(m1 - 2m2) / (m1 + m2)

Answer 2

Answer:

(2m1+3m2)/(m1+m2) m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' =V (m+m').................... Equation 1

Where m = mass of the first toy car, m' = mass of the second toy car, u = initial velocity of the of the first toy car, u' = initial velocity of the second toy car, V = common velocity of the car after collision.

make V the subject of the equation above

V = (mu+m'u')/(m+m').............. Equation 2

Let the right be positive and the left be negative

Given: m = m1, m' = m2, u = 3 m/s, u' -6 m/s (left)

Substitute into equation 2

V = [(m1×2)-(m2×-3)]/(m1+m2)

V = (2m1+3m2)/(m1+m2) m/s