Question

A 0.200 kg air-track glider moving at 1.20 m/s bumps into a 0.600 kg glider at rest.

a.) Find the total kinetic energy after collision if the collision is elastic.
b.) Find the total kinetic energy after collision if the collision is completely inelastic.

Answer 1

The other person is almost right, but I see where the numbers got fudged

Step-by-step explanation:

Answer:a) 0.144Jb) 0.036J

Explanation: A 0.200 kg air-track glider moving at 1.20 m/s bumps into a 0.600 kg glider at rest. a.) Find the total kinetic energy after collision if the collision is elastic. b.) Find the total kinetic energy after collision if the collision is completely inelastic. Given that M1 = 0.2kgM2 = 0.6kgU1 = 1.2 m/s Since both momentum and energy are conserved in elastic collisions, the total kinetic energy after collision will be1/2M1U^2 + 1/2M2U^21/2 × 0.2 × 1.2^2 + 0K.E = 0.144J

B) This is where the other person's wrong:

m1u1+m2u2=(m1+m2)V

V=(0.200kg*1.20m/s)/(0.200kg+0.600kg)=0.3m/s

KE=1/2m*v^2

KE=1/2(0.200kg+0.600kg)*(0.3m/s)^2=1/2(0.8x0.3)^2=0.036 J

Answer 2

a) 0.144J

b) 0.12J

Step-by-step explanation:

A 0.200 kg air-track glider moving at 1.20 m/s bumps into a 0.600 kg glider at rest. a.) Find the total kinetic energy after collision if the collision is elastic. b.) Find the total kinetic energy after collision if the collision is completely inelastic.

Given that

M1 = 0.2kg

M2 = 0.6kg

U1 = 1.2 m/s

Since both momentum and energy are conserved in elastic collisions, the total kinetic energy after collision will be

1/2M1U^2 + 1/2M2U^2

1/2 × 0.2 × 1.2^2 + 0

K.E = 0.144J

B) In elastic collision, only momentum is conserved

M1U1 + M2U2 = (M1 + M2)V

U2 = 0 since the object is at rest

0.2×1.2 + 0 = (0.2 + 0.6)V

0.24 = 0.8V

V = 0.24/0.8

V = 0.3 m/s

K.E = 1/2(M1+M2)V

K.E = 1/2 (0.2 + 0.6) × 0.3

K.E = 0.4 × 0.3

K.E = 0.12J