a toy car of mass 0.01 kg, gets into an elastic collision with a toy train mass 0.35 kg. if the train is at rest and the car is moving at 2 m/s initially. how fast is the train moving after the collision?
Conservation of momentum:
Mc*V1 + Mt*V2 = Mc*V3 + Mt*V4.
0.01*2 + 0.36*0 = 0.01V3 + 0.36V4.
Eq1: 0.01V3 + 0.36V4 =0.02. V3 = 2-36V4
Conservation of KE:
0.5Mc*V1^2 = 0.5Mc*V3^2 + 0.5Mt*V4^2.
0.5*0.01*2^2 = 0.005V3^2 + 0.18V4^2.
Eq2: 0.005V3^2 + 0.18V4^2 = 0.02.
Eq1 = 0.02; Eq2 = 0.02. Therefore, Eq1 = Eq2:
0.01V3 + 0.36V4 = 0.005V3^2 + 0.18V4^2.
V3 + 36V4 = 0.5V3^2 + 18V4^2
0.5V3^2-V3 + 18V4^2-36V4 = 0
0.5V3(V3-2) + 18V4(V4-2) = 0
Replace V3 with 2-36V4:
0.5(2-36V4)(-36V4) + 18V4(V4-2) = 0
(1-18V4) - 0.5(V4-2) = 0
1-18V4 - 0.5V4 + 1 = 0
18.5V4 = 2
V4 = 0.108 m/s. = Velocity of the train.
To find the velocity of the toy train after the elastic collision, we can use the law of conservation of linear momentum.
The law of conservation of linear momentum states that the total momentum before a collision is equal to the total momentum after the collision, as long as no external forces are acting on the system. Mathematically, it can be expressed as:
(m1 * v1) + (m2 * v2) = (m1 * v1') + (m2 * v2')
Where:
- m1 and v1 are the mass and velocity of the toy car before the collision.
- m2 and v2 are the mass and velocity of the toy train before the collision.
- v1' and v2' are the velocities of the toy car and toy train after the collision.
Given:
m1 (mass of toy car) = 0.01 kg
v1 (initial velocity of toy car) = 2 m/s
m2 (mass of toy train) = 0.35 kg
v2 (initial velocity of toy train) = 0 m/s (since the toy train is at rest initially)
Plugging in the values, we have:
(0.01 kg * 2 m/s) + (0.35 kg * 0 m/s) = (0.01 kg * v1') + (0.35 kg * v2')
Simplifying the equation further:
0.02 kg * m/s = 0.01 kg * v1' + 0
0.02 kg * m/s = 0.01 kg * v1'
Divide both sides by 0.01 kg:
2 m/s = v1'
Therefore, the velocity of the toy train after the collision is 2 m/s.