Question

A 2.0 kg ball and a 3.5 kg ball, each moving at 0.90 m/s, undergo a head-on collision. The lighter ball rebounds opposite its initial direction, with speed 0.90 m/s.Find the post-collision velocity of the heavier ball. Assume the initial direction of the lighter ball as positive.How much mechanical energy was lost in this collision? Express your answer in J.How much mechanical energy was lost in this collision? Express your answer as a fraction of the system's initial mechanical energy.

Answer 1

Explanation:

It is given that,

Mass of first ball, m₁ = 2 kg

Mass of other ball, m₂ = 3.5 kg

Velocities of both balls, u = 0.9 m/s

(1) The lighter ball rebounds opposite its initial direction, with speed 0.90 m/s. We need to find the final velocity of second ball. Applying the conservation of momentum as :

$2\ kg\times 0.9-3.5\ kg\times 0.9\ m/s=-2\ kg\times 0.9\ m/s+3.5v$

v is the final velocity of heavier ball.

v = 0.128 m/s

or

v = 0.13 m/s

Initial kinetic energy, $E_i=\dfrac{1}{2}\times (2\ kg+3.5\ kg)\times (0.9\ m/s)^2=2.23\ J$

Final kinetic energy, $E_f=\dfrac{1}{2}\times 2\ kg\times (0.9\ m/s)^2+\dfrac{1}{2}\times 3.5\ kg\times (0.13\ m/s)^2=0.84\ J$

Lost in kinetic energy, $\Delta KE=0.84-2.23=-1.39\ J$

Hence, this is the required solution.