A car of mass 120kg moving with a velocity of 30m/s overtake and collide with another car of mass 100kg moving with at 12m/s.After collision the two car separate with a relative velocity of 4m/s
(1)calculate the velocity each car
And
(2) The loss in kinetic after collision
m1 = 120
v1 = 30
P1 = m 1 v1 = 3600
m2 = 100
v2 = 12
P2 = m2 v2 = 1200
Total momentum before = P
P = 3600 + 1200
P = 4800
THE momentum after is THE SAME P
first mass speed after = u1
u2 = (u1+4)
so 4800 = 120 u1 + 100(u1+4)
solve for u1 and u2
Ke before = (1/2) 120 * 30^2 + (1/2) 100 * 12*2
Ke after = (1/2) 120 * u1^2 + (1/2) 100 * u2^2
subtract
Thanks man
To calculate the velocities of each car before the collision, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
1) Calculate the velocities of each car before the collision:
Let the initial velocity of the first car be v1 and the initial velocity of the second car be v2.
According to the principle of conservation of momentum:
(mass of the first car x initial velocity of the first car) + (mass of the second car x initial velocity of the second car) = (mass of the first car x final velocity of the first car) + (mass of the second car x final velocity of the second car)
(120 kg x 30 m/s) + (100 kg x 12 m/s) = (120 kg x v1) + (100 kg x v2)
(3600 kg·m/s) + (1200 kg·m/s) = 120 kg·v1 + 100 kg·v2
4800 kg·m/s = 120 kg·v1 + 100 kg·v2 ---(equation 1)
Now, after the collision, the two cars separate with a relative velocity of 4 m/s. This means the velocity of the first car after the collision is 4 m/s more than the velocity of the second car after the collision.
v1 - v2 = 4 m/s ---(equation 2)
Solving equations 1 and 2 simultaneously will give us the velocities of each car.
Using the subtraction method,
120 kg·v1 - 120 kg·v2 = 480 kg·m/s ---(equation 1) × 120 kg
120 kg·v1 - 1200 kg·v2 = 14400 kg²·m²/s²
120 kg·v1 - 1200 kg·v2 + 1200 kg·v1 - 1200 kg·v2 = 14400 kg²·m²/s² + 1200 kg·v1 + 1200 kg·v2
1320 kg·v1 - 2400 kg·v2 = 14400 kg²·m²/s² + 1200 kg·(v1 + v2) ----(equation 3)
Adding equation 2 to equation 3, we can eliminate the v2 term:
1320 kg·v1 - 2400 kg·v2 + v1 - v2 = 14400 kg²·m²/s² + 1200 kg·(v1 + v2) + 4 kg·m/s
1321 kg·v1 - 2401 kg·v2 = 14400 kg²·m²/s² + 1200 kg·(v1 + v2) + 4 kg·m/s --- (equation 4)
Substituting equation 2 into equation 4:
1321 kg·v1 - 2401 kg·(v1 - 4 m/s) = 14400 kg²·m²/s² + 1200 kg·(v1 + (v1 - 4 m/s)) + 4 kg·m/s
1321 kg·v1 - 2401 kg·v1 + 9604 kg·m/s = 14400 kg²·m²/s² + 1200 kg·(2v1 - 4 m/s) + 4 kg·m/s
-1080 kg·v1 + 9604 kg·m/s = 14400 kg²·m²/s² + 2400 kg·v1 - 4800 kg·m/s + 4 kg·m/s
-1080 kg·v1 + 9604 kg·m/s = 2400 kg·v1 + 9604 kg·m/s
-3480 kg·v1 = 0
v1 = 0 m/s
Substituting v1 = 0 into equation 2:
0 - v2 = 4 m/s
v2 = -4 m/s
Taking the absolute value of v2:
|v2| = 4 m/s
Therefore, the velocity of the first car before the collision is 0 m/s, and the velocity of the second car before the collision is 4 m/s.
2) Calculate the loss in kinetic energy after the collision:
The loss in kinetic energy is equal to the difference in kinetic energy before and after the collision.
Kinetic energy before = 1/2 x mass x velocity^2
Kinetic energy of the first car before the collision = 1/2 x 120 kg x (30 m/s)^2 = 54000 J
Kinetic energy of the second car before the collision = 1/2 x 100 kg x (12 m/s)^2 = 7200 J
Total kinetic energy before the collision = 54000 J + 7200 J = 61200 J
Kinetic energy after = 1/2 x mass x velocity^2
Kinetic energy of the first car after the collision = 1/2 x 120 kg x (0 m/s)^2 = 0 J
Kinetic energy of the second car after the collision = 1/2 x 100 kg x (4 m/s)^2 = 800 J
Total kinetic energy after the collision = 0 J + 800 J = 800 J
Loss in kinetic energy = Total kinetic energy before - Total kinetic energy after
= 61200 J - 800 J
= 60400 J
Therefore, the loss in kinetic energy after the collision is 60400 J.
To calculate the velocity of each car before the collision, we can apply the principle of conservation of momentum.
The momentum of an object is given by the mass multiplied by the velocity, represented by the equation p = m * v.
(1) Velocity of each car:
Before the collision:
Momentum of car 1 = mass of car 1 * velocity of car 1 = 120 kg * 30 m/s = 3600 kg*m/s
Momentum of car 2 = mass of car 2 * velocity of car 2 = 100 kg * 12 m/s = 1200 kg*m/s
Total momentum before the collision = momentum of car 1 + momentum of car 2
Total momentum before the collision = 3600 kg*m/s + 1200 kg*m/s = 4800 kg*m/s
After the collision:
Total momentum after the collision = momentum of car 1 + momentum of car 2
Total momentum after the collision = (mass of car 1 * velocity of car 1') + (mass of car 2 * velocity of car 2')
Total momentum after the collision = (120 kg * v1') + (100 kg * v2')
Since the two cars separate after the collision, the relative velocity between them can be used to find the velocities of each car. The relative velocity is given as 4 m/s.
Relative velocity = velocity of car 1' - velocity of car 2'
4 m/s = v1' - v2' --- (Equation 1)
Using the principle of conservation of momentum, we have:
Total momentum before the collision = Total momentum after the collision
4800 kg*m/s = (120 kg * v1') + (100 kg * v2') --- (Equation 2)
We have two equations (Equation 1 and Equation 2) with two variables (v1' and v2'). We can solve them simultaneously to find the velocities of each car.
Solving the equations, we get:
v1' = 32 m/s
v2' = 28 m/s
Therefore, the velocity of car 1 after the collision (v1') is 32 m/s, and the velocity of car 2 after the collision (v2') is 28 m/s.
(2) Loss in kinetic energy after the collision:
The initial kinetic energy of the system before the collision is given by:
KE_initial = (1/2) * (mass of car 1 * velocity of car 1^2) + (1/2) * (mass of car 2 * velocity of car 2^2)
KE_initial = (1/2) * (120 kg * 30 m/s)^2 + (1/2) * (100 kg * 12 m/s)^2
KE_initial = 540000 J + 72000 J
KE_initial = 612000 J
The final kinetic energy of the system after the collision is given by:
KE_final = (1/2) * (mass of car 1 * velocity of car 1'^2) + (1/2) * (mass of car 2 * velocity of car 2'^2)
KE_final = (1/2) * (120 kg * 32 m/s)^2 + (1/2) * (100 kg * 28 m/s)^2
KE_final = 614400 J + 392000 J
KE_final = 1002400 J
Therefore, the loss in kinetic energy after the collision is:
ΔKE = KE_initial - KE_final
ΔKE = 612000 J - 1002400 J
ΔKE = -390400 J
The loss in kinetic energy after the collision is 390400 J.