A ball with a mass of 3 kg3kg moving at 15 m/s15ms hits a still ball with a mass of 17 kg17kg. 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
Apr 6, 2017

The speed of the second ball is =2.65ms^-1=2.65ms1
The loss in kinetic energy is =277.8J=277.8J

Explanation:

We have conservation of momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2m1u1+m2u2=m1v1+m2v2

u_2=v_1=0u2=v1=0

So,

m_1u_1=m_2v_2m1u1=m2v2

v_2=m_1/m_2*u_1v2=m1m2u1

=3/17*15=2.65ms^-1=31715=2.65ms1

The kinetic energy, before the collision, is

KE_1=1/2m_1*u_1^2=1/2*3*15^2=337.5JKE1=12m1u21=123152=337.5J

The kinetic energy, after the collision, is

KE_2=1/2m_2*v_2^2=1/2*17*2.65^2=59.7JKE2=12m2v22=12172.652=59.7J

The loss in kinetis energy is

DeltaKE=KE_1-KE_2

=337.5-59.7=277.8J

Apr 6, 2017

45/17\ "m"/"s"~~2.647\ "m"/"s"

4725/17~~277.941\ "J"

Explanation:

Remember that momentum mv (m is mass and v is velocity) is always conserved.

If two objects collide, their initial momentum and their final momentum are always the same, or m_1u_1+m_2u_2=m_1v_1+m_2v_2, where m_1 and m_2 are the masses of the two objects (in this case, 3 and 17, respectively), u_1 and u_2 are their initial velocities (15 and 0, respectively), and v_1 and v_2 are their final velocities (0 and v_2, which we are trying to find).

We substitute the values given to us in the question into this formula to get 3*15+17*0=3*0+17*v_2. Solving for v_2, we get 45/17\ "m"/"s"~~2.647\ "m"/"s".

To find the amount of kinetic energy lost, we find the initial kinetic energy minus the final kinetic energy. Kinetic energy can be found using the formula (mv^2)/2, where m is mass and v is velocity.

The initial kinetic energy of the first and second ball is (3*15^2)/2=675/2\ "J" and (17*0^2)/2=0\ "J", respectively. The final kinetic energy of the first and second ball is (3*0^2)/2=0\ "J" and (17*(45/17)^2)/2=2025/34\ "J", respectively. The difference is 675/2-2025/34=4725/17~~277.941\ "J".