A 5.5 kg mess kit sliding on a frictionless surface explodes into two 2.75 kg parts, one moving at 6.0 m/s, due north, and the other at 9.0 m/s, 30° north of east. What was the original speed of the mess kit?


Somehow this physic problem seems like it should be easy, but I can't get it. I know that I use Conservation of Momentum, but I'm not sure how to go about solving for P (final). Why doesn't the below work?

P(initial) = P(final)
5.5 kg * v = 2.75 kg * 6.0 m/s + 2.75 g * 9.0 m/s * sin(30)

You probably need to find out how quickly the parts are moving away from each other to find the force of the explosion. Once you add the energy from the explosion into the right side of your equation it should work.

You need to take the momentum in the x-axis and y-axis seperately.

So, for the x-axis:

5.5*v*cos(theta) = 2.75*9.0*cos(30)

v*cos(theta) = 3.897

For the y-axis:

5.5*v*sin(theta) = 2.75*6.0 + 2.75*9.0*sin(30)

v*sin(theta) = 5.25


From this, you can pretty easily calculate v = 6.54 m/s [53.4 degrees North of East]. So the speed is just the magnitude of this or 6.54 m/s.

Scott called me in -- cause I'm in AP Physics (not bragging, just stating status quo.)

Are you in college physics? I've never heard of a type of problem like that

Actually what I said was probably wrong since the explosion would be internal to the system, and therefore already factored into the numbers. Just break the part's trajectories down into their x and y components and add those up like kyle said.

Problems like this are pretty common in calculus based AP Physics and college physics.

Oh -- we don't have Calc-based yet ... next year though!

Thanks Kyle, I didn't realize you have to compute the x and y momentum components separately. Yes, I am in college physics.

Grrr...I'm having trouble with another one. This time I can't even figure out how to start solving this problem.

In the "before" part of Figure 9-58, car A (mass 1200 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low µk of 0.14) stops them, at distances dA = 8.2 m and dB = 6.1 m.



(a) What is the speed of car A at the start of the sliding?
(b) What is the speed of car B at the start of the sliding?
(c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Any ideas guys?

Well I'm nowhere near the level of physics you're talking about... and if this is really obvious sorry... but you'd be looking at the law where something that exerts a force on something, will receive an opposite force back. So shouldn't you have some form of formula or something for that?

That's all I've got, sorry

Ummm I'm pretty elementary in physics and don't know conservation of momentum yet, but using common sense, assign the acceleration of car B (= dv/dt) and apply newton's second law ma = [summation of the] forces (net force), use the distances and the coefficient of friction along with the fact that kinetic friction = µk(normal force = mg) to solve and stuff

[edit: ok I don't know what I'm talking about. You'd have to incorporate something else to be able to effectively solve.]

Gah, you know the distance (x) and the deleration the from friction. So you have acceleration (or in this case, deceleration)and distance. Velocity shouldn't be hard to find...

So yeah, take F=ma, multiply it by mew, that is friction force, get the accel from that, then get velocity/.

Take V= v(zero) + at... or something... been so long... just solve for V! Ah so, since vfinal = 0, -At=V(initial) that should be it!

Some one tell me if I am correct!

Force of friction equals normal force times mu, and is only force acting on the cars after the collision.

F=ma
mg*mu=ma
g*mu=a

Vf^2 = Vi^2 + 2ad
Vi=sqrt(-2ad)

Once you have the velocities immediately after the collision, apply conservation of momentum.

I got 4.74, 4.09, and 8.15 m/s.

The two above post just stated what I said in a much more intelligent manner

except I don't know conservation of momentum

Thank you very much guys. I seem to have forgotten about the a = mu * g or else I might had figured it out.

There's one more problem that I am unsure of and I'm down to two submissions.

Two 2.2 kg bodies, A and B, collide. The velocities before the collision are vA = (14i + 20j) m/s and vB = (-15i + 5.0j) m/s. After the collision, v'A = (-3.0i + 22j) m/s.

a. Final velocity of B is (2.0i + 3.0j) m/s
b. What is the change in the kinetic energy (including sign?)

For b I just use Change in Kinetic = Kinetic (final) - Kinetic (initial) right? Does this mean I should add A and B's kinetic energy together for the combined kinetic (final) and kinetic (initial) before subtracting?

Okay I tried that and it didn't seem to work unless I punched in the wrong numbers...