Question

8. (9 pts.) A 4.00-kg ball, moving to the right at a velocity of +2.00 mis on a frictionless table, collides head-on with a stationary 6.50-kg ball. Find the final velocities of the balls if the collision is completely inelastic (the balls stick together). 9. (12 pts.) A 2.5-kg ball and a 5.0-kg ball have an elastic collision. Before the collision, the 2.5-kg ball was at rest and the other ball had a speed of 3.5 m/s. (a) What is the velocity of the 2.5 kg and 5.0 kg balls after the collision? (b) What is the kinetic energy of the 2.5-kg ball after the collision?

Answer 1

Given is:-

m1=2.5kg u1=0 m/s

m2=5kg u2= -3.5 m/s

we assume that 2.5kg ball is on left side and 5kg ball is on right side(we are taking left direction as negative and right as positive)

In elastic collsion we know that the momentum is conserved, thus applying conservation of momentum

$m_1u_1 + m_2u_2 = m_1v_1+m_2v_2$ eq-1

Because collision is elastic therefore kinetic energy before and after the collision will remain same

$\frac12 m_1u_1^2 + \frac12 m_2u_2^2 = \frac12 m_1v_1^2 + \frac12 m_2v_2^2$ eq-2

from both equations we get

$v_1 = \frac{m_1-m_2}{m_1+m_2}u_1 + \frac{2m_2}{m_1+m_2}u_2$

and

$v_2 = \frac{2m_1u_1}{m_1+m_2} - \frac{m_1-m_2}{m_1+m_2}u_2$

by putting values in above equations, we get

$v_1 = \frac{2.5-5}{2.5+5} \times 0 + \frac{2 \times 5}{2.5+5} \times -3.5$

$v_1 = -4.67 m/s$

and

$v_2 = 0 - \frac{2.5-5}{7.5} \times (-3.5) = 1.17 m/s$

Part-b

The kinetic energy of the 2.5 kg ball is

$K.E. = \frac12 (2.5) (-4.67)^2 = 27.26 J$