A 170g cue ball is moving at 3.0 m/s when it strikes a 160g stationary numbered ball (object ball). After the collision the balls move off as shown. Determine the speed of the cue ball and the numbered (object) ball after the collision.
The diagram illustrates the cue ball moving 60’ N of E. The object ball after the collision is 30' s of E.
you have several equations you can use.
Momenum in the original direction
Momentum in the 90 degrees direction
Conservation of energy.
Those three equations will do it.
hI bobpursley,
tHANK YOu for your prompt response. Is it possible for you to show the calculations as I've tried to answer it but to no prevail.
X 0.5
x: 0.51 kg m/s + 0= 0.17kg Vsin60' +0.16kg Vsin30'
y: 0+0= 0.17kgVcos60' +0.16kg cos30'
-0.17kgVcos60'=0.16kgcos30'
*but to no avail.
Ok, you have the momentum equations but you have failed to give each of the balls different speeds.
x:0.51 kg m/s + 0= 0.17kg Vc*sin60' +0.16kg Vo*sin30'
y: -0.17kgVc*cos60'=0.16kg*Vo*cos30'
Well, it is too easy, two equations, two unknowns. You need a teacher who made you calculate one of the angles.
Put the equations in this form...
a*Vo + b*Vc= c
d*Vo + e*Vc= d
then solve it directly ( your calculator can do it for you).
Hi,
Apologies again. But what would be the values of a, b, d and e, as well as the Vo and Vc. Our teacher had provided us with an example of one unknown value- so, we haven't been exposed to that formulae and so, am unsure as to how to use it.
To determine the speed of the cue ball and the numbered (object) ball after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum of an object is calculated by multiplying its mass by its velocity. So, the momentum (p) of an object can be written as:
p = m * v
Let's denote the velocity of the cue ball before the collision as v1, and the velocity of the numbered (object) ball before the collision as v2. Similarly, the velocities of the cue ball and the numbered (object) ball after the collision will be denoted as v1' and v2', respectively.
We can set up two equations using the principle of conservation of momentum:
1. Total momentum before the collision = Total momentum after the collision
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
2. Relationship between the velocities after the collision based on the given information:
v1' = v2' cos(30°)
v2' = v2' sin(30°)
Now, substitute the given values and solve the equations.
m1 = 170g = 0.17kg (mass of cue ball)
v1 = 3.0 m/s (velocity of cue ball before the collision)
m2 = 160g = 0.16kg (mass of numbered/object ball)
v2 = 0 m/s (velocity of numbered/object ball before the collision)
Using equation 1:
(0.17 kg * 3.0 m/s) + (0.16 kg * 0 m/s) = (0.17 kg * v1') + (0.16 kg * v2')
Simplifying:
0.51 kg m/s = 0.17 kg v1' + 0
Therefore, the velocity of the cue ball after the collision (v1') can be calculated as:
v1' = (0.51 kg m/s) / (0.17 kg) = 3.0 m/s
Using equation 2:
v1' = v2' cos(30°)
Substituting the known value for v1':
3.0 m/s = v2' cos(30°)
Rearranging:
v2' = 3.0 m/s / cos(30°) = 3.0 m/s / (√3/2) ≈ 5.2 m/s
Therefore, the velocity of the numbered (object) ball after the collision, v2', is approximately 5.2 m/s.