Question

You are playing with some toy cars. You set up a little racecar that has a mass of 120 grams to move at a velocity of 9.0 m/s due South, and a big toy truck with a mass of 350 grams at a velocity of 2.0 m/s due East. The two toys collide and stick together. What is the total kinetic energy of the toys after the collision.

Answer 1

We have : For race car:

m₁ = 120 g = 0.120 kg

u₁ = 9 m/s (south)

For truck:

m₂ = 350 g = 0.350 kg

u₂ = 2 m/s (east)

Apply law of conservation of momentum (in south) direction

m₁u₁ + m₂u₂ = (m₁+m₂) .V₁

(0.120 kg x 9 m/s) + 0 = (0.120 + 0.350 kg) x V₁

V₁ = (0.120 kg x 9 m/s) / (0.120 + 0.350 kg)

V₁ = 2.29787 m/s

Similarly, Apply law of conservation of momentum (in east direction):

m₁u₁ + m₂u₂ = (m₁+m₂) .V₂

0 + (0.350 kg x 2 m/s) = (0.120 + 0.350 kg) x V₂

V₂ = (0.350 kg x 2 m/s) / (0.120 + 0.350 kg)

V₂ = 1.48936 m/s

Hence, Net final speed is given by:

V² = V₁² + V₂²

V² = (2.29787 m/s)² + (1.48936 m/s)²

V² = 7.49841557

V = 2.738 m/s

Therefore, Final kinetic energy will be :

K.E = 1/2 (m₁+m₂) .V²

K.E = 1/2 x (0.120 + 0.350 kg) x (2.738 m/s)²

K.E = 1.762 J -------------- (**Answer**)