Particle 1 of mass 240 g and speed 3.20 m/s undergoes a one-dimensional collision with stationary particle 2 of mass 400 g.
(a) What is the magnitude of the impulse on particle 1 if the collision is elastic?
_____kg·m/s
(b) What is the magnitude of the impulse on particle 1 if the collision is completely inelastic?
_______kg·m/s
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To find the magnitude of the impulse on particle 1 in each case, we can use the principle of conservation of momentum.
(a) Elastic collision: In an elastic collision, both momentum and kinetic energy are conserved.
The momentum before the collision (momentum of particle 1) is given by:
p1_initial = mass1 * velocity1_initial
Where:
mass1 = mass of particle 1 = 240 g = 0.240 kg
velocity1_initial = initial velocity of particle 1 = 3.20 m/s
The momentum after the collision (momentum of particle 1) is given by:
p1_final = mass1 * velocity1_final
Since the collision is elastic, the final velocity can be calculated using the equation for conservation of momentum:
p1_initial = p1_final
mass1 * velocity1_initial = mass1 * velocity1_final
Solving for velocity1_final:
velocity1_final = velocity1_initial
The change in momentum (impulse) on particle 1 is given by:
Impulse = p1_final - p1_initial
Since p1_final = p1_initial, the impulse is zero:
Impulse = 0 kg·m/s
(b) Completely inelastic collision: In a completely inelastic collision, the two particles stick together after the collision and move as one body.
The momentum before the collision (momentum of particle 1) is still given by:
p1_initial = mass1 * velocity1_initial
However, to find the momentum after the collision (momentum of the combined particles), we consider the conservation of momentum:
p1_initial + p2_initial = (mass1 + mass2) * velocity_combined
Where:
p2_initial = momentum of particle 2 = 0 kg·m/s (since particle 2 is stationary)
mass2 = mass of particle 2 = 400 g = 0.400 kg
velocity_combined = velocity of the combined particles after the collision
Solving for velocity_combined:
velocity_combined = (p1_initial + p2_initial) / (mass1 + mass2)
Since p2_initial = 0 kg·m/s, this simplifies to:
velocity_combined = p1_initial / (mass1 + mass2)
The impulse on particle 1 is then given by the change in momentum:
Impulse = p1_final - p1_initial
Since the momentum after the collision is the momentum of the combined particles, we have:
p1_final = (mass1 + mass2) * velocity_combined
Substituting the value of velocity_combined:
p1_final = (mass1 + mass2) * (p1_initial / (mass1 + mass2))
Simplifying:
p1_final = p1_initial
The impulse on particle 1 is then:
Impulse = p1_final - p1_initial
Impulse = p1_initial - p1_initial
Impulse = 0 kg·m/s
Therefore, for a completely inelastic collision, the magnitude of the impulse on particle 1 is also zero:
Impulse = 0 kg·m/s
So the answers are:
(a) Magnitude of the impulse on particle 1 if the collision is elastic: 0 kg·m/s
(b) Magnitude of the impulse on particle 1 if the collision is completely inelastic: 0 kg·m/s