A 2.0-g particle moving at 8.7 m/s makes a perfectly elastic head-on collision with a resting 1.0-g object.
(a) Find the speed of each particle after the collision.
(b) Find the speed of each particle after the collision if the stationary particle has a mass of 10 g.
(c) Find the final kinetic energy of the incident 2.0-g particle in the situations described in parts (a) and (b).
In which case does the incident particle lose more kinetic energy?
Step 1
The initial velocity of the target object is zero (v2i = 0). Let m1 be the mass of the first particle, v1i its initial velocity, and v2f its final velocity. Let m2 be the mass of the target object and v2f its final velocity. From conservation of momentum before and after the collision, we have the following equation.
m1v1f + m2v2f = m1v1i + 0
For a perfectly elastic head-on collision, we have the following relationship between the objects' final velocities and initial velocities derived from conservation of momentum and conservation of kinetic energy.
v1i − v2i = − (v1f − v2f)
Because v2i = 0, we can solve for the final velocity of the second object.
v2f = v1f + v1i
Substituting equation (2) above into equation (1) and solving for v1f we have
v1f = ((m1 − m2) / (m1 + m2)) * v1i.
Substituting this result into equation (2) and simplifying, we have
v2f = (2m1 / (m1 + m2)) * v1i.
(a) If m1 = 2.0 g, m2 = 1.0 g, and v1i = 8.7 m/s, then
v1f = ((m1 − m2) / (m1 + m2)) * v1i = (1.0 g / 3.0 g) * 8.7 m/s = 2.9 m/s,
and
v2f = (2m1 / (m1 + m2)) * v1i