Block of mass sliding down incline and colliding

In summary: So, what condition does that impose on va?In summary, a block of mass 2.2 kg slides down a 30 degree incline of 3.6 m high and collides with a block of mass 7 kg at rest on a horizontal surface. The collision is elastic and friction is ignored. The speeds of the two blocks after the collision are 4.38 m/s and 4.02 m/s, respectively. The smaller mass will go 1.95 m back up the incline. The upper limit on mass m for it to rebound from M, slide up the incline, stop, slide down the incline, and collide with M again is when the velocity after the collision, va, is equal
  • #1
AnnieF
14
0

Homework Statement



A block of mass m=2.2 kg slides down a 30 degree incline which is 3.6 m high. At the bottom, it strikes a block of mass M=7 kg which is at rest on a horizontal surface. If the collision is elastic, and friction can be ignored, determine (a) the speeds of the two blocks after the collision, (b) how far back up the incline the smaller mass will go, and (c) what is the upper limit on mass m if it is to rebound from M, slide up the incline, stop, slide down the incline, and collide with M again?

Homework Equations



mgh=1/2mv^2
mghsintheta=1/2mv^2
mv1+mv2=mv1'+mv2'

The Attempt at a Solution



a) mgh=1/2mv^2
v=8.4m/s
v'a=(2.2-7/2.2+7)(8.4)=-4.38 m/s
v'b=(2*2.2/7+2.2)(8.4)= 4.02 m/s

b) mghsintheta=1/2mv^2
h=1.95 m

I am not sure if I did a and b right and I have no idea how to set c up. Any help would be appreciated!
 
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  • #2


Welcome to PF AnnieF!

Your equations and your arithmetic look good for part (a).

For part (b):

I agree that the force acting on the mass in the direction parallel to the ramp is given by mgsin(θ). However, this force has to be multiplied by the displacement of the mass in order to get the work done on the mass, which should be its change in KE. The displacement is not equal to h, so your equation is not correct.

Another way to think about it: conservation of energy says that if the mass goes up the ramp and stops, all of the mass's initial KE should be converted to potential energy, which would give us the formula:

KE = mgh

which isn't consistent with the equation you used.

For part (c):

You know that the velocity after the collision is given by:

va = [ (m-M)/(m+M) ]*v

Since there is no friction, the kinetic energy that the mass m will have after sliding up the incline, stopping, and sliding back down again will be equal to the kinetic energy it had just after the collision. (Make sure you understand why this is true, and if you don't, ask me about it!). Therefore, its speed will be the same as it was directly after the collision, only now it will be traveling towards mass M instead of away from it. So, what condition on va must be satisfied if mass m is to eventually collide with mass M, given that it is traveling in the same direction?
 
  • #3


Thank you for the correction on part b. I understand the mistake. On part C, however, I am still slightly confused. I understand the part about the kinetic energy being the same after sliding up, stopping and sliding back down as it was after the collision, but I don't understand the second part. If the velocity is the same, then shouldn't the momentum also be the same? I tried using momentum conservation, but there are too many unknowns and I am unsure of how to get the upper limit on the mass.
 
  • #4


AnnieF said:
I understand the part about the kinetic energy being the same after sliding up, stopping and sliding back down as it was after the collision, but I don't understand the second part. If the velocity is the same, then shouldn't the momentum also be the same?

The momentum would not be the same -- its magnitude would be the same, but it would have the opposite direction.

AnnieF said:
I tried using momentum conservation, but there are too many unknowns and I am unsure of how to get the upper limit on the mass.

Momentum conservation is not relevant here. You're not trying to figure out what happens after the second collision. You're trying to figure out what must be true in order to make sure that collision happens in the first place. The key to doing that is answering this question:

cepheid said:
what condition on va must be satisfied if mass m is to eventually collide with mass M, given that it is traveling in the same direction?

Can you answer this question? Hint, in order for mass m to collide with mass M on the way back, it has to catch up to it first, doesn't it?
 
  • #5


I would like to point out that the equations and calculations provided in the attempt at a solution are not entirely accurate. The equations used for calculating the speed and height of the blocks after the collision are incorrect. The correct equations to use in this scenario are the conservation of energy and conservation of momentum equations.

For part (a), the conservation of energy equation can be used to calculate the speeds of the blocks after the collision:

1/2mv^2 + 1/2Mv^2 = 1/2mv' ^2 + 1/2Mv' ^2

where v and v' are the initial and final speeds of the smaller block, and v and v' are the initial and final speeds of the larger block. Solving this equation will give the final speeds of the blocks after the collision.

For part (b), the conservation of momentum equation can be used to calculate the distance that the smaller block will travel back up the incline after the collision:

mv - Mv = mv' - Mv'

where v and v' are the initial and final speeds of the smaller block, and v and v' are the initial and final speeds of the larger block. Solving this equation will give the distance that the smaller block will travel back up the incline.

For part (c), the upper limit on the mass m can be calculated by setting the final speed of the smaller block to zero and solving for m. This will give the maximum mass of the smaller block that will allow it to rebound, slide up the incline, stop, slide down the incline, and collide with the larger block again.

I hope this helps clarify the correct approach to solving this problem. As a scientist, it is important to use accurate equations and methods when solving problems in order to obtain reliable results.
 

1. What is the equation for the kinetic energy of a block sliding down an incline?

The equation for the kinetic energy of a block sliding down an incline is given by KE = (1/2)mv^2, where m is the mass of the block and v is its velocity.

2. How does the angle of the incline affect the speed of the block?

The angle of the incline affects the speed of the block by changing the component of the force of gravity acting on the block parallel to the incline. As the angle increases, the component of the force of gravity increases, leading to a higher speed.

3. What happens to the kinetic energy of the block after it collides with a stationary object?

After colliding with a stationary object, the kinetic energy of the block is partially transferred to the object, resulting in a decrease in the block's kinetic energy. The amount of energy transferred depends on the coefficient of restitution between the two objects.

4. How does the mass of the block affect its acceleration down the incline?

The mass of the block affects its acceleration down the incline through Newton's second law, F = ma. The force of gravity acting on the block is directly proportional to its mass, so a heavier block will experience a greater force and therefore a greater acceleration down the incline.

5. What factors can affect the motion of the block as it slides down the incline?

The motion of the block as it slides down the incline can be affected by various factors, including the mass of the block, the angle of the incline, the coefficient of friction between the block and the incline, and any external forces acting on the block (such as air resistance). Additionally, the surface of the incline and the object the block collides with can also affect its motion.

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