In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The collision is cushioned by a spring (k = 1200 N/m).
Determine the velocity of each cart after the collision.
To solve this problem, we can use the principle of conservation of momentum.
Before the collision, the momentum of the first cart (Cart 1) is:
p1 = m1 * v1 = 0.60 kg * 5.0 m/s = 3.0 kg·m/s [W]
Before the collision, the momentum of the second cart (Cart 2) is:
p2 = m2 * v2 = 0.80 kg * 2.0 m/s = 1.6 kg·m/s [E]
Since this is a head-on collision, the total momentum before the collision is the sum of the individual momenta:
ptotal = p1 + p2 = 3.0 kg·m/s [W] + 1.6 kg·m/s [E] = 4.6 kg·m/s [W]
After the collision, the carts will have new velocities (v1' and v2').
Using the principle of conservation of momentum, the total momentum after the collision will also be equal to the initial total momentum:
ptotal = m1 * v1' + m2 * v2'
We also know that during the collision, the spring provides a cushioning force that will make the collision elastic. This means that kinetic energy will also be conserved.
Using the principle of conservation of kinetic energy, the initial kinetic energy is equal to the final kinetic energy:
KE_initial = KE_final
The initial kinetic energy is:
KE_initial = (1/2) * m1 * (v1^2) + (1/2) * m2 * (v2^2)
KE_initial = (1/2) * 0.60 kg * (5.0 m/s)^2 + (1/2) * 0.80 kg * (2.0 m/s)^2
KE_initial = (1/2) * 0.60 kg * 25.0 m^2/s^2 + (1/2) * 0.80 kg * 4.0 m^2/s^2
KE_initial = 37.50 J + 1.60 J
KE_initial = 39.10 J
The final kinetic energy is:
KE_final = (1/2) * m1 * (v1'^2) + (1/2) * m2 * (v2'^2)
KE_final = (1/2) * 0.60 kg * (v1'^2) + (1/2) * 0.80 kg * (v2'^2)
Since kinetic energy is conserved, we can set the initial kinetic energy equal to the final kinetic energy and solve for v1' and v2':
37.50 J + 1.60 J = (1/2) * 0.60 kg * (v1'^2) + (1/2) * 0.80 kg * (v2'^2)
Rearranging the equation, we get:
(v1'^2) + (v2'^2) = (39.10 J) / (0.60 kg) + (0.80 kg)
(v1'^2) + (v2'^2) = 65.17 m^2/s^2
Since the collision is head-on, the carts will move in opposite directions after the collision. Therefore, we can express the velocities as:
v1' = -u
v2' = u
Substituting these values into the equation, we get:
(-u)^2 + u^2 = 65.17 m^2/s^2
2u^2 = 65.17 m^2/s^2
u^2 = 65.17 m^2/s^2 / 2
u^2 = 32.585 m^2/s^2
u = sqrt(32.585 m^2/s^2)
u ≈ 5.71 m/s
Therefore, the velocity of Cart 1 after the collision (v1') is approximately -5.71 m/s [W], and the velocity of Cart 2 after the collision (v2') is approximately 5.71 m/s [E].
In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The collision is cushioned by a spring (k = 1200 N/m).
Determine the maximum compression of the spring.
To determine the maximum compression of the spring, we can use the principle of conservation of mechanical energy.
Before the collision, the total initial mechanical energy is the sum of the kinetic energy of both carts:
E_initial = (1/2) * m1 * (v1^2) + (1/2) * m2 * (v2^2)
E_initial = (1/2) * 0.60 kg * (5.0 m/s)^2 + (1/2) * 0.80 kg * (2.0 m/s)^2
E_initial = (1/2) * 0.60 kg * 25.0 m^2/s^2 + (1/2) * 0.80 kg * 4.0 m^2/s^2
E_initial = 37.50 J + 1.60 J
E_initial = 39.10 J
After the collision, the total final mechanical energy is the sum of the potential energy of the compressed spring and the kinetic energy of both carts:
E_final = (1/2) * k * (x^2) + (1/2) * m1 * (v1'^2) + (1/2) * m2 * (v2'^2)
Since this is an elastic collision, the final kinetic energy is equal to the initial kinetic energy:
E_final = (1/2) * k * (x^2) + (1/2) * m1 * (v1^2) + (1/2) * m2 * (v2^2)
E_final = (1/2) * k * (x^2) + E_initial
Setting the initial mechanical energy equal to the final mechanical energy, we get:
E_initial = E_final
39.10 J = (1/2) * k * (x^2) + E_initial
Rearranging the equation, we have:
(1/2) * k * (x^2) = E_initial - E_initial
(1/2) * k * (x^2) = 0
x^2 = 0
Since the equation has no solution when x^2 = 0, this means that the spring is not compressed after the collision.
Therefore, the maximum compression of the spring is 0 meters.
Choose one of the following problems and solve it completely:
At a road intersection, a 1200 kg car travelling southward at 18.0 m/s collides with a 975 kg car travelling eastward at 33.3 m/s. As seen in the diagram below. The cars remain together after the collision. Determine the direction and speed that the damaged cars travel after they collide (ignore friction).
To solve this problem, we can use the principle of conservation of momentum.
Before the collision, the momentum of the first car (Car 1) is:
p1 = m1 * v1 = 1200 kg * (-18.0 m/s) = -21600 kg·m/s [S]
Before the collision, the momentum of the second car (Car 2) is:
p2 = m2 * v2 = 975 kg * (33.3 m/s) = 32467.5 kg·m/s [E]
Since the cars remain together after the collision, their total momentum will be the sum of the individual momenta:
ptotal = p1 + p2 = -21600 kg·m/s [S] + 32467.5 kg·m/s [E] = 10867.5 kg·m/s [E]
After the collision, the cars will have a combined mass of:
m = m1 + m2 = 1200 kg + 975 kg = 2175 kg
Using the principle of conservation of momentum, the total momentum after the collision will also be equal to the initial total momentum:
ptotal' = m * v'
10867.5 kg·m/s [E] = 2175 kg * v'
v' = 10867.5 kg·m/s [E] / 2175 kg
v' = 5.0 m/s [E]
Therefore, after the collision, the damaged cars will travel together at a speed of 5.0 m/s [E].
Jogger 1 is travelling east at 6.5 m/s and has a mass of 82 kg. Jogger 2 is travelling north at 5.8 m/s and has a mass of 54.5 kg. One of the joggers has their head down and doesn’t see the other, resulting in a right-angle collision. Both joggers are locked together after the collision. Determine the direction and speed of the two joggers after the collision.
To solve this problem, we can use the principle of conservation of momentum.
Before the collision, the momentum of Jogger 1 (J1) is:
p1 = m1 * v1 = 82 kg * (6.5 m/s) = 533 kg·m/s [E]
Before the collision, the momentum of Jogger 2 (J2) is:
p2 = m2 * v2 = 54.5 kg * (5.8 m/s) = 316.1 kg·m/s [N]
Since the joggers are locked together after the collision, their total momentum will be the sum of the individual momenta:
ptotal = p1 + p2 = 533 kg·m/s [E] + 316.1 kg·m/s [N]
To determine the direction and speed of the joggers after the collision, we can calculate the resultant momentum using the Pythagorean theorem:
ptotal = sqrt((ptotal)^2) = sqrt((533)^2 + (316.1)^2) = sqrt(284089 + 99872.21) = sqrt(383961.21) = 619.98 kg·m/s
The direction of the resultant momentum can be found using the inverse tangent function:
θ = atan(p2 / p1) = atan(316.1 kg·m/s / 533 kg·m/s) = atan(0.592)
Therefore, the angle that the resultant momentum makes with the positive x-axis is approximately 29.39 degrees.
To determine the direction and speed of each jogger after the collision, we can use the concept of vector resolution. Since the joggers are locked together and move as one unit, they will have the same speed after the collision.
Using the resultant momentum and the cosine function, we can determine the speed (v') of the joggers after the collision:
v' = ptotal / mtotal = 619.98 kg·m/s / (82 kg + 54.5 kg) = 6.00 m/s
Using the speed and the sine and cosine functions, we can determine the velocity components (vx' and vy') of the joggers after the collision:
vx' = v' * cos(θ) = 6.00 m/s * cos(29.39 degrees) = 5.23 m/s [E]
vy' = v' * sin(θ) = 6.00 m/s * sin(29.39 degrees) = 2.94 m/s [N]
Therefore, after the collision, the joggers will move together with a speed of 6.00 m/s in a direction approximately 31.59 degrees north of east.