A ball with a mass of #14 kg# moving at #2 m/s# hits a still ball with a mass of #20 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?

Question

A ball with a mass of #14 kg# moving at #2 m/s# hits a still ball with a mass of #20 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?

Adrian Ayala · Accepted Answer

undefined After the collision, the second ball will be moving at approximately 1.4 m/s. The kinetic energy lost as heat in the collision is approximately 56 J.

Amelia Adams · Answer

Please see the explanation below

We have conservation of momentum #m_1u_1+m_2u_2=m_1v_1+m_2v_2# The mass the first ball is #m_1=14kg# The velocity of the first ball before the collision is #u_1=2ms^-1# The mass of the second ball is #m_2=20kg# The velocity of the second ball before the collision is #u_2=0ms^-1# The velocity of the first ball after the collision is #v_1=0ms^-1# Therefore, #14*2+20*0=14*0+20*v_2# #20v_2=28# #v_2=28/20=1.4ms^-1# The velocity of the second ball after the collision is #v_2=1.4ms^-1# The loss in kinetic energy is #DeltaKE=KE_i-KE_f# #=1/2*14*2^2-1/2*20*1.4^2# #=28-19.6# #=8.4J# The kinetic energy lost in the collision is #=8.4J#

Elena Albarran · Answer

undefined To determine the velocity of the second ball after the collision, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.

$ \text{Initial momentum} = \text{Final momentum} $

$ m_1 \times v_1 + m_2 \times v_2 = 0 + m_2 \times v_f $

$ 14 \times 2 + 20 \times 0 = 0 + 20 \times v_f $

$ 28 = 20 \times v_f $

$ v_f = \frac{28}{20} = 1.4 \, \text{m/s} $

So, the second ball is moving at $ 1.4 \, \text{m/s} $ after the collision.

To find the amount of kinetic energy lost as heat in the collision, we can calculate the initial kinetic energy and the final kinetic energy.

Initial kinetic energy ($ KE_i $) = $ \frac{1}{2} \times m_1 \times v_1^2 $

$ KE_i = \frac{1}{2} \times 14 \times (2)^2 = 28 \, \text{J} $

Final kinetic energy ($ KE_f $) = $ \frac{1}{2} \times m_2 \times v_f^2 $

$ KE_f = \frac{1}{2} \times 20 \times (1.4)^2 = 19.6 \, \text{J} $

The kinetic energy lost as heat in the collision is:

$ \text{Kinetic energy lost} = KE_i - KE_f $

$ \text{Kinetic energy lost} = 28 \, \text{J} - 19.6 \, \text{J} = 8.4 \, \text{J} $

So, $ 8.4 \, \text{J} $ of kinetic energy was lost as heat in the collision.