Question

Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piece has a mass of 300 grams and is moving to the right at a speed of 1 m/s. The other piece has mass 600 grams and is moving to the left at a speed of 0.75 m/s. What fraction of the total initial kinetic energy is lost during the collision?

Answer 1

Answer:

0.961

Explanation:

m1 = 300 g = 0.3 kg

u1 = 1 m/s

m2 = 600 g = 0.6 kg

u2 = - 0.75 m/s

Let after collision they move together with velocity v.

By using the conservation of linear momentum

Total momentum before collision = Total momentum after collision

m1 x u1 + m2 x u2 = (m1 + m2) v

0.3 x 1 - 0.6 x 0.75 = (0.3 + 0.6) v

0.3 - 0.45 = 0.9 v

v = - 0.166 m/s

Total initial Kinetic energy

$K_{i}=0.5m_{1}u_{1}^{2}+0.5m_{1}u_{1}^{2}$

$K_{i}=0.5\times 0.3\times 1\times 1+0.5 \times 0.6 \times 0.75 \times 0.75$

$K_{i}=0.31875 J$

Total final Kinetic energy

$K_{f}=0.5\left ( m_{1}+m_{2} \right )v^{2}$

$K_{f}=0.5\times 0.9 \times 0.166 \times 0.166$

$K_{f}=0.0124 J$

fraction of kinetic energy lost

$\frac{K_{i}-K_{f}}{K_{f}}=\frac{0.31875-0.0124}{0.31875}=0.961$