A ball with a mass of #8 kg# moving at #7 m/s# hits a still ball with a mass of #16 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 #8 kg# moving at #7 m/s# hits a still ball with a mass of #16 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?

Adam Arnold · Accepted Answer

undefined To find the velocity of the second ball after collision, we use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.

$m_1 \times v_1 + m_2 \times v_2 = m_1 \times u_1 + m_2 \times u_2$

$8 \times 7 + 16 \times 0 = 8 \times 0 + 16 \times v_2$

$56 = 16 \times v_2$

$v_2 = \frac{56}{16} = 3.5 \, \text{m/s}$

To find the kinetic energy lost as heat in the collision, we first calculate the initial kinetic energy and the final kinetic energy:

Initial kinetic energy = $ \frac{1}{2} \times m_1 \times u_1^2 + \frac{1}{2} \times m_2 \times u_2^2 $

Final kinetic energy = $ \frac{1}{2} \times m_1 \times 0^2 + \frac{1}{2} \times m_2 \times v_2^2 $

Kinetic energy lost as heat = Initial kinetic energy - Final kinetic energy

Adrian Ayala · Answer

#"velocity of the second object after collision : "v_2^'=3.5" "m/s#

#"the lost kinetic energy is 100 Joules"# #"Momentum and Kinetic energy before collision :"# #".........................................................................."# #vec P_1:"momentum of the first object before collision"# #m_1:"mass of the first object"# #v_1:"velocity of the first object before collision"# #vec P_1=m_1*v_1# #vec P_1=8*7=56 " "kg*m/s# #vec P_2:"momentum of the second object before collision"# #m_2:"mass of the second object"# #v_2:"velocity of the second object before collision"# #vec P_2=m_2*v_2# #vec P_2=16*0=0# #Sigma vec P_b:"the vectorial sum of the momentums before collision"# #Sigma vec P_b=vec P_1+vec P_2# #Sigma vec P_b=56+0=56 " "kg*m/s# #Sigma E_k:"Total kinetic energy"# #Sigma E_k=1/2*m_1*v_1^2+0" (the second object hasn't kinetic energy)"# #Sigma E_k=1/2*8*7^2=4*49=196" Joules"# #"Momentum and Kinetic energy after collision :"# #".........................................................................."# #P_1^'=m_1*v_1^'# #"So "v_1^'=0" ; "P_1^'=8*0=0# #P_2^'=m_2*v_2^'# # Sigma vec P_a=P_1^'+P_2^'" total momentum after..."# #Sigma vec P_a=0+16*v_2^'# #Sigma vec P_b=Sigma vec P_a# #56=16*v_2^'# #v_2^'=56/16=3.5 " "m/s# #E_k=1/2*m_2*v_2^('2)+0=1/2*16*(3.5)^2=8*12.25# #E_k=98" Joules"# #Delta E_k=196-96# #Delta E_k=100" Joules ( lost kinetic energy)" #