A ball with a mass of #4# #kg# and velocity of #5# #ms^-1# collides with a second ball with a mass of #6# #kg# and velocity of #-1# #ms^-1#. If #20%# of the kinetic energy is lost, what are the final velocities of the balls?

Question

Lily Adams · Accepted Answer

Momentum is conserved. Kinetic energy is not, but in this case we know that final #E_k# is 0.8 times initial #E_k#.

The final velocity of the #4# #kg# ball is #-2.38# #ms^-1# and the final velocity of the #6# #kg# ball is #2.73# #ms^-1#.

Let's call the #4# #kg# ball Ball 1 and the #6# #kg# ball Ball 2, just for the convenience of subscripts... Prior to the collision: vigor: #p=m_1v_1+m_2v_2=4xx5+6xx(-1)=14# #kgms^-1# Energy in motion: #E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2xx4xx5^2+1/2xx6xx(-1)^2# #=50+3=53# #J# Following the collision: vigor: #p=m_1v_1+m_2v_2=4v_1+6v_2# Energy in motion: #E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2xx4xxv_1^2+1/2xx6xxv_2^2# We know that momentum is conserved, so the momentum after the collision will be #14# #kgms^-1#, as it was before the collision. 20% of the kinetic energy is lost to other forms of energy like heat and sound, so 80% will remain: #0.8xx53=33.9# #J# We are left with two equations involving two unknowns: Equation 1: #4v_1+6v_2=14# Equation 2: #2v_1^2+3v_2^2=33.9# Rearranging Equation 1 to make #v_1# the subject: #v_1=(14-6v_2)/4# Putting this in place of Equation 2: #2((14-6v_2)/4)^2+3v_2^2=33.9# It's just algebra, so I'll leave you to solve this equation, which involves solving a quadratic equation. The solution is that #v_2=2.73# or #1.75# #ms^-1# Substituting this back into Equation 1 and solving we find that #v_1=-2.38# or #3.5# #ms^-1# respectively. Assume that the "+" direction is from left to right and the "-" direction is from right to left in order to determine which set of answers we want. Initially, the #4# #kg# ball was moving left-to-right and the #6# #kg# ball was moving right-to-left. After the collision, we can imagine that they have both bounced in the opposite direction: the #4# #kg# ball is now traveling right-to-left and has a negative velocity, the #6# #kg# ball has a positive velocity. Physically speaking, the other set of answers is nonsensical because it assumes that the left-hand ball is moving more quickly to the right than the right-hand ball.

Benjamin Asiedu · Answer

#4kg " ball"=-1ms^-1 and 6kg " ball"=3ms^-1#
#4kg " ball"=1.52ms^-1 and 6kg " ball"=1.32ms^-1# We notice that velocity of one ball is given as #5ms^-1# and of the other is given as #-1ms^-1#. This implies that both the balls are moving in a direction opposite to each other. both the balls have been given. Let #x# direction be defined by the positive velocity. Before the collision Momentum, in #x# direction only #p_i=4xx5+6xx(-1)=14kgms^-1# Kinetic energy #KE_i=1/2xx4xx5^2+1/2xx6xx(-1)^2# #=50+3=53J# After the collision Let #v_(1x) and v_(2x)# be #x# components of respective velocities of #4kg and 6kg# balls and #v_(1y) and v_(2y)# be #y# components of their respective velocities. Momentum: #p_(fx)=4v_(1x)+6v_(2x)# and #p_(fy)=4v_(1y)+6v_(2y)# Since momentum is conserved, we have #14=4v_(1x)+6v_(2x)# .....(1) #0=4v_(1y)+6v_(2y)# .....(2) Kinetic energy: #KE_f=1/2xx 4(v_(1x)^2+v_(1y)^2)+1/2 xx6(v_(2x)^2+v_(2y)^2)# Given is that #20%# kinetic energy is lost. #:. KE_f=0.8xx53=42.4J# This gives us #42.4=2(v_(1x)^2+v_(1y)^2)+3(v_(2x)^2+v_(2y)^2)# .....(3) Thus we have three equations in four unknowns. To find a solution lets assume that even after collision the balls move along the #x# direction only. Implies that equation (2) vanishes. And we have two equations to solve for two variables. #4v_(1x)+6v_(2x)=14# ......(4) #2v_(1x)^2+3v_(2x)^2=42.4# ....(5) Dropping the suffix #x# we get #4v_1+6v_2=14# ......(6) #2v_1^2+3v_2^2=42.4# ....(7) Substituting value of #v_1# from (6) in (7) #v_1=(14-6v_2)/4# #=>2((14-6v_2)/4)^2+3v_2^2=42.4# #=>(196-168v_2+36v_2^2)+3v_2^2=42.4# #=>39v_2^2-168v_2+153.6=0# #=>13v_2^2-56v_2+51.2=0# Solving the quadratic #v_2=(-(-56)+-sqrt((-56)^2+4xx13xx51.2))/(2xx13)# #v_2=(56+-sqrt(3136-4xx13xx51.2))/(26)# #v_2=(56+-21.8)/(26)# #v_2=3, 1.32# From (6) corresponding values of #v_1=-1, 1.52#

Carson Alexander · Answer

undefined To solve for the final velocities of the balls after the collision, we can use the principle of conservation of momentum and the concept of kinetic energy loss.

Let $ v_1 $ and $ v_2 $ be the final velocities of the first and second balls, respectively.

According to the conservation of momentum:

$ m_1 \cdot v_{1i} + m_2 \cdot v_{2i} = m_1 \cdot v_1 + m_2 \cdot v_2 $

where $ m_1 $ and $ m_2 $ are the masses of the first and second balls, and $ v_{1i} $ and $ v_{2i} $ are their initial velocities.

Given:
$ m_1 = 4 $ kg, $ v_{1i} = 5 $ m/s,
$ m_2 = 6 $ kg, $ v_{2i} = -1 $ m/s.

Applying the conservation of momentum:

$ 4 \cdot 5 + 6 \cdot (-1) = 4 \cdot v_1 + 6 \cdot v_2 $

$ 20 - 6 = 4v_1 + 6v_2 $

$ 14 = 4v_1 + 6v_2 $ ...(1)

Now, to consider the kinetic energy loss, let $ KE_i $ be the initial kinetic energy and $ KE_f $ be the final kinetic energy.

Given that 20% of kinetic energy is lost, $ KE_f = 0.8 \times KE_i $.

$ KE_i = \frac{1}{2} m_1 v_{1i}^2 + \frac{1}{2} m_2 v_{2i}^2 $

$ KE_f = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 $

Substituting the given values:

$ 0.8 \times (\frac{1}{2} \cdot 4 \cdot 5^2 + \frac{1}{2} \cdot 6 \cdot (-1)^2) = \frac{1}{2} \cdot 4 \cdot v_1^2 + \frac{1}{2} \cdot 6 \cdot v_2^2 $

$ 0.8 \times (50 + 3) = 2v_1^2 + 3v_2^2 $

$ 0.8 \times 53 = 2v_1^2 + 3v_2^2 $

$ 42.4 = 2v_1^2 + 3v_2^2 $ ...(2)

Now, we have two equations (equation 1 and equation 2) with two unknowns $ v_1 $ and $ v_2 $. We can solve them simultaneously to find the final velocities of the balls.