A ball of mass 0.640 kg moving east (+x direction) with a speed of 3.10 m/s collides head-on with a 0.780 kg ball at rest. If the collision is perfectly elastic, what will be the speed and direction of each ball after the collision?
1. ball originally at rest
2. ball originally moving east
im alittle confused about this problem.
if you could give more details and give me the equations for me to start. that would be very helpful. much appreciated
Perfectly elastic collision assumes conservation of energy. Momentum is conserved for both elastic and inelastic collisions.
Let 1 and 2 denote each of the balls (1 moving, 2 initially stationary).
m1 and m2 their masses
v11 and v21 their initial velocities (v21=0)
v12 and v22 their final velocities.
Conservation of Kinetic energy equation:
(1/2)m1 v11² + (1/2)m2 v21²
= (1/2)m1 v12² + (1/2)m2 v22² ....(1)
Conservation of Momentum equation:
m1 v11 + m2 v21 = m1 v12 + m2 v22 ...(2)
All variables are known except for v12 and v22.
Solve for v12 and v22.
wow, much help. thank you very much but i do have a quick question,
v21² AND v22² are both in the same equation, are you saying that im supposed to put v21² AND v22² on the same side? like this....
(1/2)m1 v11² + (1/2)m2 (1/2)m1 v12² + (1/2)m2
= v21² v22² ??????
(1/2)m1 v11² + (1/2)m2 v21²
This is the left hand side of the equation and represents the sum of kinetic energies of the two balls before the impact.
The right hand side of the equation:
(1/2)m1 v12² + (1/2)m2 v22²
represents the total kinetic energies after the elastic impact.
They are equated because energy is supposed to be conserved during an elastic impact.
The confusion probably arose because the equal sign was at the beginning of the next line.
By the way, you will probably find that v12 is negative, which physically means that ball 1 bumped backwards while pushing ball 2 forward.
Given:
M1 = 0.640kg, V1 = 3.10 m/s.
M2 = 0.780kg, V2 = 0.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
0.64*3.10 + 0.78*0 = 0.64*V3 + 0.78*V4.
Eq1: 0.64*V3 + 0.78*V4 = 1.984.
Conservation of KE Eq:
a. V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (3.10(0.64-0.78) + 1.56*0)/(1.42) = -0.306 m/s,Left = Velocity of M1(0.64kg) after the collision.
b. In Eq1, replace V3 with -0.306 and solve for V4.
Sure! To solve this problem, we can use the principles of conservation of momentum and kinetic energy. The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. The conservation of kinetic energy states that the total kinetic energy before the collision is equal to the total kinetic energy after the collision in an elastic collision.
Let's analyze each scenario separately:
1. Ball originally at rest:
In this case, before the collision, the 0.640 kg ball is moving in the +x direction with a speed of 3.10 m/s, and the 0.780 kg ball is at rest (0 m/s).
To find the final speed and direction of each ball after the collision, we can use the following equations:
Conservation of momentum:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
Conservation of kinetic energy:
(1/2) * m1 * (v1_initial)^2 + (1/2) * m2 * (v2_initial)^2 = (1/2) * m1 * (v1_final)^2 + (1/2) * m2 * (v2_final)^2
Plugging in the given values:
m1 = 0.640 kg (mass of the moving ball)
v1_initial = 3.10 m/s (initial speed of the moving ball)
m2 = 0.780 kg (mass of the ball at rest)
v2_initial = 0 m/s (initial speed of the ball at rest)
Solving these equations will give us v1_final and v2_final (final speeds of each ball), which will allow us to determine the direction.
2. Ball originally moving east:
In this case, before the collision, both balls are moving in the +x direction. The 0.640 kg ball is moving with a speed of 3.10 m/s, and the 0.780 kg ball is at rest (0 m/s).
To solve this scenario, we will use similar equations as above, but with different initial velocities.
I hope this clarifies the problem and provides the necessary equations to start solving it!