An air-track cart of mass 0.10 is moving with a speed 1.2 to the right when it collides with a cart of mass 0.24 that is at rest. Each cart has a wad of putty on its bumper, and hence they stick together as a result of their collision. Suppose the average contact force between the carts is 1.3N during the collision. Cart 1 has an acceleration of 13 m/s^2. Cart 2 is accelerating at 5.4 m/s^2.
How long does it take for both carts to have the same speed? (Once the carts have the same speed the collision is over and the contact force vanishes.)
What is the final speed of the carts, ?
The mass of cart 1 is F/a1 = 0.1 kg
The mass of cart 2 is F/a2 = 0.241 kg
The final speed Vf of both carts at the end of the collision is given by momentum conservation:
0.341 Vf = 0.1*1.2
Vf = 0.352 m/s
The time required for cart 1 to go from 1.2 m/s to 0.341 m/s while decelerating at 13 m/s^2 is
0.0682 s.
To find the time it takes for both carts to have the same speed, we can use the equation of motion:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
1) For Cart 1:
Given:
Mass of Cart 1 (m1) = 0.10 kg
Acceleration of Cart 1 (a1) = 13 m/s^2
Initial velocity of Cart 1 (u1) = 1.2 m/s
Time taken for Cart 1 (t1) = ?
Using the equation of motion for Cart 1:
v1 = u1 + a1 * t1
Since we want to find the time taken for Cart 1 to have the same speed as Cart 2, the final velocity of Cart 1 will be the same as the final velocity of Cart 2 (v2). Therefore, we can substitute v1 = v2.
v2 = u2 + a2 * t2 ... (Equation 1)
2) For Cart 2:
Given:
Mass of Cart 2 (m2) = 0.24 kg
Acceleration of Cart 2 (a2) = 5.4 m/s^2
Initial velocity of Cart 2 (u2) = 0 m/s
Time taken for Cart 2 (t2) =?
Using the equation of motion for Cart 2:
v2 = u2 + a2 * t2
Substituting v1 = v2 from Equation 1:
u2 + a2 * t2 = u2 + a1 * t1
Simplifying:
a2 * t2 - a1 * t1 = 0
Dividing both sides by the acceleration difference:
(t2 - t1) = 0
Since the contact force vanishes when both carts have the same speed, the time taken for Cart 1 (t1) and Cart 2 (t2) will be equal.
Therefore, the time taken for both carts to have the same speed is zero.
Now let's find the final speed of the carts:
Using the equation of motion for Cart 1:
v1 = u1 + a1 * t
Substituting v1 = v2:
v2 = u1 + a1 * t
Substituting the given values:
v2 = 1.2 + 13 * 0
Simplifying:
v2 = 1.2 m/s
Hence, the final speed of the carts is 1.2 m/s.
To find the time it takes for both carts to have the same speed, we can use the concept of acceleration and the constant contact force between the carts during the collision.
First, let's calculate the acceleration of the carts using Newton's second law, F = ma, where F is the net force exerted on the cart and m is the mass of the cart.
For Cart 1:
Mass (m1) = 0.10 kg
Acceleration (a1) = 13 m/s^2
Therefore, the net force acting on Cart 1 (F1) can be calculated as:
F1 = m1 * a1
F1 = 0.10 kg * 13 m/s^2
F1 = 1.3 N
For Cart 2:
Mass (m2) = 0.24 kg
Acceleration (a2) = 5.4 m/s^2
Similarly, the net force acting on Cart 2 (F2) can be calculated as:
F2 = m2 * a2
F2 = 0.24 kg * 5.4 m/s^2
F2 = 1.3 N
Since the average contact force during the collision is the same for both carts (1.3 N), we can equate the two forces:
F1 = F2
m1 * a1 = m2 * a2
Now, to find the time it takes for both carts to have the same speed, we can use the relationship between acceleration, time, and change in velocity:
a = (v_f - v_i) / t
Since the initial velocity for Cart 1 is 1.2 m/s and Cart 2 is at rest, the equation can be written as:
a1 * t = -a2 * t
Simplifying the equation, we have:
(a1 + a2) * t = 0
Since a1 = 13 m/s^2 and a2 = 5.4 m/s^2:
(13 + 5.4) * t = 0
18.4 * t = 0
Since the acceleration sum is zero, the time it takes for both carts to have the same speed is 0 seconds. This implies that they have the same speed instantaneously once the contact force vanishes.
To find the final speed of the carts after 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 since there is no external force acting on the system:
m1 * v1i + m2 * v2i = (m1 + m2) * vf
Substituting the given values:
(0.10 kg * 1.2 m/s) + (0.24 kg * 0 m/s) = (0.10 kg + 0.24 kg) * vf
0.12 kg m/s = 0.34 kg * vf
vf = 0.12 kg m/s / 0.34 kg
vf ≈ 0.35 m/s
Therefore, the final speed of the carts after the collision is approximately 0.35 m/s