A snooker ball with a mass of 0.3kg moving at 2.5m/s collides with another ball of 0.2kg mass and moving in the same direction with a speed of 1.20m/s . Assuming the collision is elastic, calculate the speed and direction of each ball after collision.
same momentum before and after
0.3*2.5 + 0.2*1.2 = 0.3 u + 0.2 v
but elastic so same kinetic energy too
(1/2)*.3*2.5^2+(1/2)*.2*1.2^2
=(1/2)*.3*u^2+(1/2)*.2*v^2
momentum is conserved
and with an elastic collision, energy is conserved
capital letters - larger ball
small letters - smaller ball
MV1 + mv1 = MV2 + mv2
... (.3 * 2.5) + (.2 * 1.2) = .3 V2 + .2v2
1/2 MV1^2 + 1/2 mv1^2 = 1/2 MV2^2 + 1/2 mv2^2
... (.15 * 6.25) + (.144) = .15 V2^2 + .1 v2^2
two equations...two unknowns...algebra time
both balls continue moving in the same direction
Given:
M1 = 0.3kg, V1 = 2.5 m/s.
M2 = 0.2kg, V2 = 1.2 m/s.
Momentum before = Momentum after:
M1*V1 = M2*V2 = M1*V3 + M2*V4.
0.3*2.5 + 0.2*1.2 = 0.3*V3 + 0.2*V4,
Eq1: 0.3*V3 + 0.2*V4 = 0.99.
Kinetic Energy Eq.:
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3 = (2.5(0.3-0.2) + 0.4*1.2)/(0.3+0.2),
V3 = (0.25 + 0.48)/0.5 = 1.46 m/s.
In Eq1, replace V3 with 1.46 and solve for V4.
Well, the snooker ball and the other ball really had a smashing encounter! Let's calculate the speeds and directions after this jolly ol' elastic collision.
First, we need to conserve momentum. The total momentum before the collision is equal to the total momentum after the collision. So, we have:
(0.3 kg)(2.5 m/s) + (0.2 kg)(1.20 m/s) = (0.3 kg)(v1) + (0.2 kg)(v2)
Now, let's find the final speeds of the balls (v1 and v2). Solving the equation above, we get:
0.75 + 0.24 = 0.3v1 + 0.2v2
1.49 = 0.3v1 + 0.2v2
To further solve this mystery, we need to make use of the conservation of kinetic energy in elastic collisions. The sum of the kinetic energies before the collision must be equal to the sum of the kinetic energies after the collision. So, we have:
(1/2)(0.3 kg)(2.5 m/s)^2 + (1/2)(0.2 kg)(1.20 m/s)^2 = (1/2)(0.3 kg)(v1)^2 + (1/2)(0.2 kg)(v2)^2
0.9375 + 0.144 = 0.15(v1)^2 + 0.02(v2)^2
1.0815 = 0.15(v1)^2 + 0.02(v2)^2
Now, to solve the system of equations simultaneously, we can substitute the value of v2 from the first equation into the second equation:
1.0815 = 0.15(v1)^2 + 0.02(1.49 - 0.3v1)
1.0815 = 0.15(v1)^2 + 0.0298 - 0.006v1
1.0517 = 0.15(v1)^2 - 0.006v1
Alright, now we bring in our trusty pal, the quadratic formula:
v1 = [-(-0.006) ± √((-0.006)^2 - 4(0.15)(-1.0517))] / (2(0.15))
v1 = [0.006 ± √(0.000036 + 0.633)) / 0.3
After a delightful calculation, we find two possible values for v1: 0.2908 m/s and -10.0908 m/s. Since we assume both balls are moving in the same direction, we reject the negative value. We're not dealing with time machines here!
Now, we can use the first equation to find v2:
0.75 + 0.24 = 0.3(0.2908) + 0.2v2
0.99 = 0.08724 + 0.2v2
0.90276 = 0.2v2
v2 = 4.5138 m/s
So, after this elastic collision, the snooker ball will move forward with a speed of approximately 0.2908 m/s, and the other ball will continue its journey with a speed of approximately 4.5138 m/s, both in the same direction.
Hope this playful explanation didn't make your head spin too much!
To solve this problem, we need to apply the principles of conservation of momentum and conservation of kinetic energy.
First, let's calculate the initial momentum of both balls before the collision.
Momentum (p) is given by the product of the mass (m) and velocity (v):
p = m * v
For the first ball with a mass of 0.3 kg and a velocity of 2.5 m/s:
p1 = 0.3 kg * 2.5 m/s = 0.75 kg·m/s
For the second ball with a mass of 0.2 kg and a velocity of 1.20 m/s:
p2 = 0.2 kg * 1.20 m/s = 0.24 kg·m/s
Next, let's calculate the total momentum before the collision:
p_initial = p1 + p2 = 0.75 kg·m/s + 0.24 kg·m/s = 0.99 kg·m/s
According to the conservation of momentum principle, the total momentum before the collision should be equal to the total momentum after the collision.
Now, let's consider the speed and direction of each ball after the collision. Since the collision is elastic, both the momentum and the kinetic energy will be conserved.
Let's assume that the first ball moves in the positive x-direction, and the second ball moves in the same direction as well.
Let the final velocities of the first and second balls be v1f and v2f, respectively.
Using the conservation of momentum, we have:
p1_initial + p2_initial = p1_final + p2_final
Substituting the values:
0.75 kg·m/s + 0.24 kg·m/s = 0.3 kg * v1f + 0.2 kg * v2f
Using the conservation of kinetic energy, we have:
(1/2) * m1 * v1_initial^2 + (1/2) * m2 * v2_initial^2 = (1/2) * m1 * v1_final^2 + (1/2) * m2 * v2_final^2
Substituting the values:
(1/2) * 0.3 kg * (2.5 m/s)^2 + (1/2) * 0.2 kg * (1.20 m/s)^2 = (1/2) * 0.3 kg * (v1f)^2 + (1/2) * 0.2 kg * (v2f)^2
Solving these two equations will give us the final velocities v1f and v2f, which will determine the speed and direction of each ball after collision.