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UTF-8 U+6211 U+662F wrote:

Justin72835 wrote:

UTF-8 U+6211 U+662F wrote:

Answer? (Click to Show Hidden Text)

1) kMv/(M+m) 2) 2kMv/(M+m)

You're definitely on the right track (I see that you already found the final velocities of the objects after the collision). Using the final velocity, you can find the kinetic energy of the object after the collision and set it equal to 1/2kx^2 and solve for x. This works because maximum compression occurs when the objects kinetic energy has been converted completely into potential energy.

For those who want it (Click to Show Hidden Text)

[math]x_{inelastic}=Mv\sqrt{\frac{1}{k(M+m)}}[/math]

[math]x_{elastic}=\frac{2Mv}{M+m}\sqrt{\frac{m}{k}}[/math]

Sorry, this question is from a long time ago, but I want to check if I'm doing it right (Click to Show Hidden Text)

1)
[math]Fd = \frac12(M+m)v_{final}^2[/math], where [math]v_{final}[/math] represents the speed of the cube after the collision happens.
[math](kd)d = \frac12(M+m)v_{final}^2[/math]
[math]kd^2 = \frac12(M+m)v_{final}^2[/math]
[math]Mv = (M+m)v_{final}[/math]
[math]v_{final} = \frac{Mv}{M+m}[/math]
[math]kd^2 = \frac12(M+m)\left(\frac{Mv}{M+m}\right)^2[/math]
[math]d^2 = \frac1{2k}(M+m)\left(\frac{Mv}{M+m}\right)^2[/math]
[math]d^2 = \frac1{2k}\frac{(Mv)^2}{M+m}[/math]
[math]d^2 = \frac{(Mv)^2}{2k(M+m)}[/math]
[math]d = \sqrt{\frac{(Mv)^2}{2k(M+m)}}[/math]
[math]d = Mv\sqrt{\frac1{2k(M+m)}}[/math]
But this is slightly different from the posted answer, so where did I go wrong?

2)
[math]kd^2 = \frac12mv_{final}^2[/math]
[math]Mv = Mv_1 + mv_2[/math]
[math]v = v_2 - v_1[/math]
[math]v_1 = v_2 - v[/math]
[math]Mv = M(v_2 - v) + mv_2[/math]
[math]Mv = Mv_2 - Mv + mv_2[/math]
[math]2Mv = v_2(M+m)[/math]
[math]v_2 = \frac{2Mv}{M+m}[/math]
[math]kd^2 = \frac12m(\frac{2Mv}{M+m})^2[/math]
[math]d^2 = \frac1{2k}m(\frac{2Mv}{M+m})^2[/math]
[math]d^2 = \frac{m(2Mv)^2}{2k(M+m)^2}[/math]
[math]d = \frac{2Mv}{M+m}\sqrt{\frac{m}{2k}}[/math]
Again, this answer is slightly different posted answer, so where did I go wrong?

Edit: Sorry about the double post too.

You went wrong in assuming that the force applied by the spring remains constant. 

The work done by the spring isn't kd^2, it's 1/2kd^2 (potential energy of a spring). I had a really hard time understanding this when I first began learning about energy and started wondering where the '1/2' term in elastic potential energy (1/2kd^2) and kinetic energy (1/2mv^2) came from. For the first one, you can think of it like this:

When the spring is uncompressed, it applies a force of 0; when it is fully compressed, it applies a force of kd. Thus the average force applied by the spring is 1/2kd. Plugging this into W=Fd gives W=1/2kd^2. You can also prove this using calculus but this is a far more intuitive way of thinking about it.

The '1/2' term cancels out with the '1/2' from the kinetic energy equation, which is why my answer differs by a factor of root(1/2).

Justin72835 wrote: A company is experimenting with different bowling ball designs. A hollow bowling ball and a uniform bowling ball (same mass and radius) are positioned at the top of an inclined plane with a height of 5 meters. If they roll without slipping down the inclined plane, what is its translational velocity upon reaching the bottom of the plane of both balls?

Note: it may seem like you need more information to solve this but if done correctly then a bunch of terms will cancel out (kind of like how mass always seems to cancel out in energy equations).

Riptide wrote:

Justin72835 wrote: A company is experimenting with different bowling ball designs. A hollow bowling ball and a uniform bowling ball (same mass and radius) are positioned at the top of an inclined plane with a height of 5 meters. If they roll without slipping down the inclined plane, what is its translational velocity upon reaching the bottom of the plane of both balls?

Note: it may seem like you need more information to solve this but if done correctly then a bunch of terms will cancel out (kind of like how mass always seems to cancel out in energy equations).

8.37 m/s for the uniform ball and 7.67 m/s for the hollow ball.

Riptide wrote:A x directed force is applied to an object with the equation F_x = 5+.23x N. Find the work done by the force as the object moves from (0,0) to (7,0).

Assuming coordinate system is in meters, [math]\int^{7 m}_{0 m} (5 + .23x) N dx = 40.635 J[/math]?

Riptide wrote:A x directed force is applied to an object with the equation F_x = 5+.23x N. Find the work done by the force as the object moves from (0,0) to (7,0).

Awesome calc question!

[math]W=\int_{0}^{7}Fdx=\int_{0}^{7}(5+.23x)dx=40.64 N[/math]

UTF-8 U+6211 U+662F wrote:

Riptide wrote:A x directed force is applied to an object with the equation F_x = 5+.23x N. Find the work done by the force as the object moves from (0,0) to (7,0).

Answer (Click to Show Hidden Text)

Assuming coordinate system is in meters, [math]\int^{7 m}_{0 m} (5 + .23x) N dx = 40.635 J[/math]?

UTF-8 U+6211 U+662F wrote:All right, next question! Ignore the completely unrealistic values.
An 80.0 kg astronaut experiences 0.100 'g's and throws a wrench that weighs 10.00 pound (at sea level on Earth) with an angle of elevation above his center of mass of 30.0 degrees. The wrench flies away from his hand at a speed of 100.0 m/s.
a) How fast does he move after one second? Two seconds? Assume that gravity pulls on him downward with a constant force and that he will never hit ground. Also assume that the time spent throwing the wrench is negligible.
b) What is the impulse of the wrench on the astronaut? Assume constant acceleration on the wrench.
c) Suppose the mass of the planet the astronaut is above is 1.00 * 10^23 kg. How high does the astronaut need to be above the center of mass of the planet to experience the gravity he experiences?

Pretty fun problem tbh.

A) 10 lb = 4.54 kg. Using conservation of momentum, we find that the speed of the astronaut immediately after throwing the wrench is 5.67 m/s. This velocity is divided up into two components, x and y: vx = 4.91 m/s and vy = 2.83 m/s. 

After one second, vy increases by 0.98 m/s, so using Pythagorean Theorem, we find that his total velocity is [b]6.21 m/s[/b].

After two seconds, vy increases by 1.96 m/s, so using the same procedure, we find that his total velocity is [b]6.86 m/s[/b].

B) Impulse is just change in momentum, which in this case is [b]452.6 kg m/s[/b].

C) a = Gm/r^2. Substituting a for 0.98 m/s^2 and m for 1e23 kg, the radius comes out to be [b]2610 km[/b].

Justin72835 wrote:

UTF-8 U+6211 U+662F wrote:All right, next question! Ignore the completely unrealistic values.
An 80.0 kg astronaut experiences 0.100 'g's and throws a wrench that weighs 10.00 pound (at sea level on Earth) with an angle of elevation above his center of mass of 30.0 degrees. The wrench flies away from his hand at a speed of 100.0 m/s.
a) How fast does he move after one second? Two seconds? Assume that gravity pulls on him downward with a constant force and that he will never hit ground. Also assume that the time spent throwing the wrench is negligible.
b) What is the impulse of the wrench on the astronaut? Assume constant acceleration on the wrench.
c) Suppose the mass of the planet the astronaut is above is 1.00 * 10^23 kg. How high does the astronaut need to be above the center of mass of the planet to experience the gravity he experiences?

Answer (Click to Show Hidden Text)

Pretty fun problem tbh.

A) 10 lb = 4.54 kg. Using conservation of momentum, we find that the speed of the astronaut immediately after throwing the wrench is 5.67 m/s. This velocity is divided up into two components, x and y: vx = 4.91 m/s and vy = 2.83 m/s. 

After one second, vy increases by 0.98 m/s, so using Pythagorean Theorem, we find that his total velocity is [b]6.21 m/s[/b].

After two seconds, vy increases by 1.96 m/s, so using the same procedure, we find that his total velocity is [b]6.86 m/s[/b].

B) Impulse is just change in momentum, which in this case is [b]452.6 kg m/s[/b].

C) a = Gm/r^2. Substituting a for 0.98 m/s^2 and m for 1e23 kg, the radius comes out to be [b]2601 km[/b].