Question

Two clay balls collide head-on in a perfectly inelastic collision. The first ball has a mass of 0.500 kg and an initial velocity of 4.0 m/s to the right. The mass of the second ball is 0.250 kg, and it has an initial velocity of 3.00 m/s to the left. What is the final velocity of the combined ball of clay after the collision

Answer 1

Answer:

$v'=1.667\ m.s^{-1}$  in the direction of the first body

Explanation:

Given:

• mass of the first ball, $m_1=0.5\ kg$

• velocity of the first ball, $v_1=4\ m.s^{-1}$

• mass of the second ball, $m_2=0.25\ kg$

• velocity of the second ball, $v_2=3\ m.s^{-1}$

Now for the head-on inelastic collision:

$m_1.v_1-m_2.v_2=(m_1+m_2)v'$

(since the bodies combine after an inelastic collision.)

$0.5\times 4-0.25\times 3=(0.5+0.25)\times v'$

$v'=1.667\ m.s^{-1}$ in the direction of the first body. (Since the net result is positive as assumed in the equation for the first body)

Answer 2

Answer:

Explanation:

mass of first ball, m1 = 0.5 kg

mass of second ball, m2 = 0.25 kg

initial velocity of first ball, u1 = 4 m/s

initial velocity of the second ball, u2 = - 3 m/s

Let v be the final velocity of the balls after collision

By use of conservation of momentum

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

0.5 x 4 - 0.25 x 3 = ( 0.5 + 0.25) x v

2 - 0.75 = 0.75 v

v = 1.67 m/s towards right.