Tarzan and Jane(collision in one dimension problem)

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In summary, Jane swings into the path of a pack of stampeding elephants and uses a rope vine to haul Tarzan to safety. If Jane's mass is 56 kg, and Tarzan's mass is 91 kg, to what height above the ground will the pair swing after she rescues him?
  • #1
gongshow29
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Homework Statement


Tarzan is in the path of a pack of stampeding elephants when Jane swings into the rescue on a rope vine, hauling him off to safety. The length of the vine is 27 m, and Jane starts her swing with the rope horizontal. If Jane's mass is 56 kg, and Tarzan's mass is 91 kg, to what height above the ground will the pair swing after she rescues him? (Assume the rope is vertical when she grabs him.)

Homework Equations


(1/2)mv^2=KE
mgh=PE

The Attempt at a Solution


I know that this is an inelastic collision, but I do not know how to set up the problem, I was thinking along the lines of:
(1/2)mv^2=mgh after finding the velocity of Jane, but I don't know how to do that. Help?
 
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  • #2
Here's a hint: What is the difference between Jane's height at the beginning of the swing and her height at the lowest point of her swing (when we assume she picks up Tarzan)?

Also, after she picks up Tarzan, treat the two as one combined mass.

Basically, split the problem into a before-pickup and after-pickup. You have the right idea in considering mgh and (1/2)mv².
 
  • #3
Ok, so I tried to set it up as follows:
(56)(9.8)(27)=(1/2)(56)v^2
so for the velocity I got 23.00434742.
then i plugged this velocity into the follwing:
(1/2)(56+91)v^2=mgh
and solved for h, and I ended up getting 27 again which is incorrect, which part of my method is incorrect?
 
  • #4
The mistake is in what you are equating. We need not find Jane's speed at the lowest point; rather, we must only find the energy she has gained. This energy then goes into raising the combined masses of her and Tarzan to a new height. Let J = mass of Jane and T = mass of Tarzan.

E(before) = E(after)
J(9.8)(27)=(J+T)(9.8)h

... and solve for the height.

I believe you made the mistake in the following step:
(1/2)(56+91)v^2=mgh

You see, by adding the mass of Tarzan where you did, you added mass while keeping the velocity constant, which equals energy from nowhere. (Impossible.)
 
  • #5
I used your method above and came out with something like this:
(56kg)(9.8)(27m)=(56kg+91kg)(9.8)(h),
solving for h I get 10.285 and it is still telling me that the answer is incorrect.
This is the hint that it gives me,
"Jane's collision with Tarzan is a perfectly inelastic collision. Find her speed just before she grabs Tarzan from conservation of energy and their speed just after she grabs him from conservation of momentum. Their kinetic energy just after their collision will be transformed into gravitational potential energy when they have reached their greatest height."
 
  • #6
Ok, then try this:

Use mgh = 1/2mv^2 to find Jane's velocity at the lowest point in her swing.

Now use conservation of momentum:

m(initial)v(initial) = m(final)v(final)

m(initial) is just jane's mass and v(initial) is the velocity you found.
m(final) is the combined mass of both jane and tarzan.

Solve for v(final) and plug that number into 1/2mv^2 to get energy.
Now equate to mgh again and find height.
 
  • #7
Ok that ended up working, thank you very much for the help, I guess on the first try when I calcluated the velocity i plugged in the inital velocity of jane instead of the final velocity.
 

Related to Tarzan and Jane(collision in one dimension problem)

1. What is the "Tarzan and Jane" collision in one dimension problem?

The "Tarzan and Jane" collision in one dimension problem is a physics problem that involves calculating the collision between two objects moving in a straight line, with one object starting from rest while the other is in motion. It is named after the famous characters Tarzan and Jane, who are often used as examples in this type of problem.

2. How is the "Tarzan and Jane" collision in one dimension problem solved?

The problem is solved using the principles of conservation of momentum and conservation of kinetic energy. The initial and final velocities of the two objects are used to calculate their masses and the distance between them at the point of collision. These values are then used to determine the final velocities of the objects after the collision.

3. What are the assumptions made in the "Tarzan and Jane" collision in one dimension problem?

The problem assumes that the collision between the two objects is perfectly elastic, meaning there is no loss of kinetic energy during the collision. It also assumes that there are no external forces acting on the objects, such as friction or air resistance. Additionally, it assumes that the objects are point masses with no rotational motion.

4. Can the "Tarzan and Jane" collision in one dimension problem be applied to real-life situations?

Yes, the principles used in this problem can be applied to real-life situations involving collisions between objects moving in a straight line. However, in real-life scenarios, there are often external forces and other factors that can affect the outcome of the collision.

5. What are some examples of situations where the "Tarzan and Jane" collision in one dimension problem can be applied?

This problem can be applied to situations such as a car collision, where two cars are moving towards each other and collide, or a sports scenario where a ball is thrown towards a stationary player. It can also be used in scenarios involving billiard balls, air hockey pucks, or any other objects that collide in a straight line.

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