2 insects collide in midair at the top of their trajectories,hold each other.Calc the change in kinetic energy

Question

insect A has mass 250g and is moving south at 10.0 cm/s , while insect B is 120g initially moving north at 60.0 cm/s . (Answer must be in Joules)

Answer 1

Robert is correct that you use conservation of momentum to determine the velocity of the resulting 370g of insects.

Answer 2

Change the grams to kilograms
250 g = 0.250 kg
120 g = 0.120 kg

Change the velocity to m/sec
10.0 cm/sec = 0.100 m/sec
60.0 cm/sec = 0.600 m/sec

First, find the velocity of the insects after the collision using the momentum equation

Mv + mv = (M + m)v
0.250(0.100) + 0.120(-0.600) = (0.250 + 0.120)v
v = _____ m/sec
[The reason why 0.600 is negative is that you are dealing with velocity, not speed.]

Change in KE = KE before collision - KE after collison.
Use the equation KE = 1/2 mv^2

KE south moving insect = 1/2 (0.250)(0.100)^2
KE north moving insect = 1/2 (0.120)(-0.600)^2
Add the two numbers together: This is the KE before the collision.

Find the KE of the insects after the collision = 1/2 (0.250 + 0.120)v^2
[the v is the number you found in the momentum equation]

Now subtract the two numbers.--The answer you get will be the change in KE, in Joules.

Answer 3

Find the combined kinetic energy and net momentum of the insects before the collision. Use conservation of momentum to find the velocity of the insects after collision. From that velocity calculate the final kinetic energy, and subtract the intitial energy from that.