A ball with a mass of 4 kg moving at 3 m/s hits a still ball with a mass of 26 kg. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

2 Answers

6/13=0.461\ \text{m/s} & loss of K.E. =15.231\ J

Explanation:

Ball of mass m_1=4kg moving with u_1=3\ m/s strikes another ball of mass m_2=26\ kg at rest u_2=0. After collision, first ball stops v_1=0 & second ball moves with the velocity v_2 then by law of conservation of momentum we have

m_1u_1+m_2u_2=m_1v_1+m_2v_2

4(3)+26(0)=4(0)+26v_2

v_2=12/26

=6/13\ \text{m/s}

Loss of kinetic energy

=\text{initial K.E.}-\text{final K.E.}

=1/2m_1u_1^2-1/2m_2v_2^2

=1/2(4)(3)^2-1/2(26)(6/13)^2

=15.231\ J

Jul 28, 2018

The speed is =0.46ms^-1 and the loss in kinetic energy is =15.23J

Explanation:

There is conservation of momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

The mass m_1=4kg

The velocity u_1=3ms^-1

The mass m_2=26kg

The velocity u_2=0ms^-1

The velocity v_1=0ms^-1

Therefore,

4*3+26*0=4*0+26*v_2

v_2=12/26=0.46ms^-1

The loss in kinetic energy is

DeltaKE=KE_1-KE_2

=1/2m_1u_1^2-1/2m_2v_2^2

=1/2*4*3^2-1/2*26*0.46^2

=15.23J