Suppose that the graph of force vs. time for a very unusual spring has only straight lines. The cart hits the spring at t = 3.00 s, so force F = 0 up to that point. Then, after t = 3.00 s, the force increases linearly until it reaches a maximum at Fmax = 6.74 N at time t = 3.55 s. Then the spring pushes the cart away, and the force decreases linearly until it goes back to F = 0 at time t = 3.85 s.
a) Find the magnitude of the impulse exerted by the spring on the cart.
b) Find the speed of the cart just as it leaves the spring.
HINT: Assume the cart is at rest when the force exerted by the spring is a maximum.
To find the magnitude of the impulse exerted by the spring on the cart, we can use the formula:
Impulse = Force * Time
In this case, the force is changing over time, so we need to break the calculation into two parts: the force when it is increasing, and the force when it is decreasing.
a) Finding the impulse when force is increasing:
First, we need to find the area under the force vs. time graph during the increasing part. Since the graph is a straight line, we can calculate the area of the triangle formed by the base and height.
Area = (1/2) * base * height
The base of the triangle is the time interval when the force is increasing: t = 3.55s - 3.00s = 0.55s
The height of the triangle is the maximum force: Fmax = 6.74N
Now we can calculate the area:
Area = (1/2) * 0.55s * 6.74N = 1.8485 Ns
This is the impulse during the increasing part of the force.
b) Finding the impulse when force is decreasing:
Similarly, we need to find the area under the force vs. time graph during the decreasing part.
The base of the triangle is the time interval when the force is decreasing: t = 3.85s - 3.55s = 0.30s
The height of the triangle is also the maximum force: Fmax = 6.74N
Now we can calculate the area:
Area = (1/2) * 0.30s * 6.74N = 1.011 Ns
This is the impulse during the decreasing part of the force.
To find the total impulse exerted by the spring on the cart, we add the impulses during the increasing and decreasing parts:
Total Impulse = Impulse (increasing) + Impulse (decreasing)
= 1.8485 Ns + 1.011 Ns
= 2.8595 Ns
Therefore, the magnitude of the impulse exerted by the spring on the cart is 2.8595 Ns.
c) Finding the speed of the cart just as it leaves the spring:
The impulse exerted by a force on an object is equal to the change in momentum of the object. Therefore, we can use the formula:
Impulse = Change in momentum
The momentum of an object is given by:
Momentum = Mass * Velocity
In this case, we want to find the speed of the cart, which is the magnitude of the velocity.
Now we can use the formula:
Impulse = Change in momentum
2.8595 Ns = (Mass of the cart) * (Final Velocity - Initial Velocity)
Since the initial velocity of the cart is zero (at rest), we can simplify the equation:
2.8595 Ns = (Mass of the cart) * (Final Velocity)
We need more information about the cart, specifically its mass, in order to solve for the final velocity.
To find the magnitude of the impulse exerted by the spring on the cart, we can use the formula for impulse:
Impulse = ∫F dt
Since the force is linear, we can break the integral into two parts: the increase in force and the decrease in force.
The first part of the integral represents the increase in force from F = 0 to Fmax = 6.74 N:
Impulse1 = ∫(0 to Fmax) F dt
Since the force is increasing linearly, we can find the average force during this interval by taking half of the maximum force:
Average Force1 = (0 + Fmax)/2 = Fmax/2
The time interval for the increase in force is from t = 3.00 s to t = 3.55 s:
Time interval1 = 3.55 s - 3.00 s = 0.55 s
Therefore, the impulse for the increase in force is:
Impulse1 = Average Force1 * Time interval1
Plugging in the values:
Impulse1 = (6.74 N/2) * (0.55 s) = 1.855 N·s
The second part of the integral represents the decrease in force from Fmax = 6.74 N to F = 0:
Impulse2 = ∫(Fmax to 0) F dt
Again, since the force is decreasing linearly, we can find the average force during this interval by taking half of the maximum force:
Average Force2 = (Fmax + 0)/2 = Fmax/2
The time interval for the decrease in force is from t = 3.55 s to t = 3.85 s:
Time interval2 = 3.85 s - 3.55 s = 0.30 s
Therefore, the impulse for the decrease in force is:
Impulse2 = Average Force2 * Time interval2
Plugging in the values:
Impulse2 = (6.74 N/2) * (0.30 s) = 1.005 N·s
Finally, to find the total impulse:
Total Impulse = Impulse1 + Impulse2
Total Impulse = 1.855 N·s + 1.005 N·s = 2.860 N·s
Therefore, the magnitude of the impulse exerted by the spring on the cart is 2.860 N·s.
To find the speed of the cart just as it leaves the spring, we can use the impulse-momentum theorem:
Impulse = Change in momentum
The impulse exerted by the spring is equal to the change in momentum of the cart. Since the cart starts at rest, the change in momentum is equal to the final momentum.
Therefore, we can write:
Impulse = mv - 0
where m is the mass of the cart and v is its final velocity.
Since the impulse is given as 2.860 N·s and the mass is not provided in the question, we cannot directly calculate the velocity.