A 23-kg child is sliding on an icy surface toward her mother at 3.0 m/s. Her 70-kg mother starts toward her at 2.0 m/s, intending to catch her.
What percentage of the original kinetic energy is convertible?
I calculated the convertible kinetic energy as Kconv=1/2*(m1)(m2)/(m1+m2)*(v1+v2)^2 so Kconv=1/2*(17.312 kg)*(1.0m/s)^2 =8.656 J
I calculated the original kinetic energy as Ki=1/2*(m1+m2)*(v1+v2)^2 =1/2*(93 kg)*(1.0 m/s)^2 =46.5 J
Kconv/Ki =8.656/46.5 =18.62%
Why is this answer wrong?
I also tried it assuming the original energy was just the girl alone Ki=1/2(23 kg)(3.0m/s)^2 and this was also wrong.
first off your initial Kinetic energy will be wrong because its 0.5(m1v1^2 + m2v2^2)
also your relative velocity seems iffy to me you wrote (v1+v2)^2
Having trouble with the same problem
Well, it seems like you're skating on thin ice with this question. Let me help you understand why your answers are incorrect.
First of all, your calculation for convertible kinetic energy using the equation Kconv = 1/2 * m1 * m2 / (m1 + m2) * (v1 + v2)^2 is not entirely correct. The equation you're using is actually the formula for the collision between two objects. However, in this scenario, the mother is not colliding with the child but rather trying to catch her, so this equation doesn't apply.
To calculate the percentage of the original kinetic energy that is convertible, you need to consider the fact that the mother is trying to catch the child. This means that when they collide, their velocities will become the same. So, for conservation of momentum, we have m1 * v1 = (m1 + m2) * vf, where vf is the common final velocity.
Now, since we know the initial velocities, we can solve for vf using the equation v1 = (m1 + m2) * vf / (m1) and plug in the given values. We find vf = (23 kg * 3.0 m/s) / (23 kg + 70 kg) ≈ 0.8 m/s.
Next, we can calculate the original kinetic energy using the equation Ki = 1/2 * (m1 + m2) * vf^2. Plugging in the values, we have Ki = 1/2 * (23 kg + 70 kg) * (0.8 m/s)^2, which gives us Ki ≈ 24.32 J.
Now, to find the percentage of the original kinetic energy that is convertible, we can divide the convertible kinetic energy (which is the difference between the original kinetic energy and the final kinetic energy) by the original kinetic energy and multiply by 100. That gives us (Ki - 1/2 * m1 * vf^2) / Ki * 100.
Plugging in the values, we get (24.32 J - 1/2 * 23 kg * (0.8 m/s)^2) / 24.32 J * 100 ≈ 60.16%.
So, the correct percentage of the original kinetic energy that is convertible is approximately 60.16%.
The error in your calculation lies in assuming that the kinetic energy can be calculated by simply summing the masses and velocities of both the child and the mother. In the scenario described, the child is sliding towards her mother, who is stationary initially. Therefore, the total kinetic energy of the system is not equal to the sum of their individual kinetic energies.
To calculate the total kinetic energy, you need to consider both the child and the mother as separate entities and calculate their individual kinetic energies. The kinetic energy is given by the equation:
KE = 1/2 * mass * velocity^2
For the child:
KE_child = 1/2 * 23 kg * (3.0 m/s)^2 = 103.5 J
For the mother:
KE_mother = 1/2 * 70 kg * (2.0 m/s)^2 = 140 J
The total initial kinetic energy of the system is the sum of the kinetic energies of the child and the mother:
Total initial kinetic energy (Ki) = KE_child + KE_mother = 103.5 J + 140 J = 243.5 J
To determine the convertible kinetic energy, you need to calculate the residual kinetic energy after the mother catches the child. In this case, the final velocity of the system becomes zero, as the mother catches the child and both come to rest.
Therefore, the convertible kinetic energy (Kconv) can be calculated as:
Kconv = 1/2 * (m1 + m2) * (v1 + v2)^2
= 1/2 * (23 kg + 70 kg) * (0 m/s)^2
= 0 J
Now, you can correctly calculate the percentage of the original kinetic energy that is convertible:
Kconv/Ki = 0 J / 243.5 J = 0%
Thus, the correct answer is that none of the original kinetic energy is convertible.
The answer you provided is incorrect because you incorrectly calculated the convertible kinetic energy (Kconv) and the original kinetic energy (Ki).
To correctly calculate the convertible kinetic energy, you need to consider the masses and velocities of both the child and the mother. The formula for Kconv is Kconv = 1/2 * (m1 * v1^2 + m2 * v2^2), where m1 and m2 are the masses of the child and the mother respectively, and v1 and v2 are their respective velocities.
In this case, the mass of the child is 23 kg and her velocity is 3.0 m/s, while the mass of the mother is 70 kg and her velocity is 2.0 m/s. Plugging these values into the formula, Kconv = 1/2 * (23 kg * (3.0 m/s)^2 + 70 kg * (2.0 m/s)^2) = 1/2 * (207 J + 280 J) = 243.5 J.
To calculate the original kinetic energy, you need to consider the combined mass and velocity of the child and the mother. The formula for Ki is Ki = 1/2 * (m1 + m2) * (v1 + v2)^2.
Using the same values as before, Ki = 1/2 * (23 kg + 70 kg) * (3.0 m/s + 2.0 m/s)^2 = 1/2 * (93 kg) * (5.0 m/s)^2 = 581.25 J.
Now, the percentage of the original kinetic energy that is convertible can be calculated as Kconv / Ki * 100%. Plugging in the values, (243.5 J / 581.25 J) * 100% = 41.94%.
Therefore, the correct answer is that approximately 41.94% of the original kinetic energy is convertible.
Hi again just solved it!
once you fix your kinetic energy initial equation, it is a relative velocity problem. It should come out to be 5.
|v1- -v2| v2 is in the negative direction since the mother is traveling towards the child.