Momentum, kinematics, and dymanics ?

[songoku]

Homework Statement

A block of wood of mass m1 = 20 kg is placed on a cart of mass 2 = 15 kg, which is stationary on a frictionless surface. The coefficient of kinetic friction between the wood and the cart is 0.6 A cannon ball of mass m3 = 5 kg which is fired from a cannon, begins to penetrate the wood at t = 0. If the velocity of the cannon ball just before it hits the wood is 20 m/s and assuming that the cannon ball experiences a constant resistive force of 8000 N throughout the penetration.
a. Find the penetration depth of the ball and the time it takes to reach this depth
b. Find out the distance that the wood moves on the cart.

Homework Equations

Momentum, kinematics, and dynamics

The Attempt at a Solution

a.
m₁u₁ + m₃u₃ = (m₁+m₃)v

v = 4 ms^-1 (the speed of the ball after impact)

The deceleration of the ball = \frac{8000}{5}=1600~ms^{-2}

v²=u²-2as ; u = 4 ms^-1 , v = 0
s = 1.25 x 10^-3 m (the depth of penetration)


b.
delta Ek = friction*s (take wood as consideration)

0.5*20*16 = 200*s

s = 0.8 m


Do I get it right?

Thanks

 

[turin]

There are three objects: block, cart, and ball. Each one has a state of motion. The block interacts with the ball and the cart. All three have finite inertia, and there is no other interaction with the outside universe. Fortunately, this is only a 1-D problem.

So, your first equation, m1u1 + m3u3 = (m1+m3)v, is wrong, because it neglects the interaction (friction) between the block and the cart. That is, the cart will also impart impulse to the block (actually, it will do so in the opposite direction to the ball).

Assuming that the block remains on the cart in the final state, consider all three objects as a closed system, with initial state momentum still equal to the initial momentum of the ball.

 

[songoku]

Hi turin

I still don't know how to set up the equation. Consider all the objects as one system, the initial momentum will be the same as initial momentum of the ball.
How about the final momentum? I think because all the object move after the impact, they all contribute to the final momentum, and maybe the friction also? Will the speed of the three objects be the same after impact because we consider them as one system?

Thanks

 

[turin]

I guess "system" is somewhat arcane (and probably a bit misleading, since we don't want to bother with accounting for heat). Sorry, I will rephrase:

The objects must share the momentum among themselves, and the total of the momentum of the three objects must always add up to the same amount. This is due to the assumption that, in the effective 1-D of the problem, there are no forces (effectively) acting from "outside".¹ The friction, or any force whatsoever, between any of the three objects does not alter this situation (Newton's third law).

In the final state, you should assume that all three objects move as one. So, it is sort of like an inelastic collision.

Hmm... I just realized another complication, but you are probably supposed to ignore it; I don't know. I will mention it just in case:

To lowest approximation, even if the ball impacts the block perfectly horizontally, it should experience, IN ADDITION TO "the constant resistive force", a support force from the block directed upwards in order to counteract its weight, assuming that it continues to penetrate in a perfectly horizontal direction so that "the constant resistive force" acts in a perfectly horizontal direction. This in turn would translate to an additional normal force between the block and the cart. I wouldn't mention this if m_ball << m_block. However, m_ball = (1/3)m_block, so the effect is significant. Like I said, I don't know if you're supposed to ignore this or not.



¹ So here I am ignoring, for instance, the change in momentum due to gravity on the ball, and only considering the x-direction, say. However, you still need to use normal force, in the y-direction, say, to calculate friction.

 

[songoku]

Hi turin

So the equation for conservation of momentum should be :
m₁u₁ + m₂u₂+m₃u₃ = (m₁+m₂+m₃)v ??

Then, find the deceleration which is 1600 ms^-2(like in my first post) and use v²=u²-2as to find the depth of penetration (v is obtained from the above equation for momentum). Also use kinematic formula to find the time taken. (v = u - at)

For (b), use delta Ek = friction*s to find the distance. I also don't know I should ignore the fact you mention or not, but I think I will consider it in my work. So, the normal force for finding the friction will be : μk (m₁+m₃)g

Am I on the right track?

Thanks

 

[turin]

For (b), use delta Ek = friction*s to find the distance.

I don't think that will work. Ek is the kinetic energy? Of what? The block is initially not moving. Then, it is set in motion due to impulse from the ball. That is, it gains kinetic energy from something like F_res⋅s. However, it sumulatneously loses energy through friction (something like what you wrote). Then, somehow, the ball and block reach the same speed, and the block and cart reach the same speed.

I also don't know I should ignore the fact you mention or not, but I think I will consider it in my work. So, the normal force for finding the friction will be : μk (m₁+m₃)g

I believe you mean that to be the friction force already.

 

[songoku]

Hi turin. Sorry I take long time to reply

I don't think that will work. Ek is the kinetic energy? Of what? The block is initially not moving. Then, it is set in motion due to impulse from the ball. That is, it gains kinetic energy from something like F_res⋅s. However, it sumulatneously loses energy through friction (something like what you wrote). Then, somehow, the ball and block reach the same speed, and the block and cart reach the same speed.

Ek is kinetic energy of the block. The block moves after the impact, then it stops after moving certain distance so I consider that in the formula. The block moves with velocity obtained from the equation of momentum then stops because of the friction. The change in kinetic energy is the work done by the friction. That's my thought but I think it's wrong according to you. Maybe you can point out my mistake.

I believe you mean that to be the friction force already.

Oh yes. That's the friction force, not the normal force.

Thanks a lot

 

[turin]

You have two basic "errors". (At least, you are not being specific enough for me to tell whether or not you have the correct idea.)

1) What is the final speed of the block? (i.e., does it lose all of its kinetic energy?)

2) What will you consider as the initial kinetic energy of the block? Or, better yet, can you plot the kinetic energy w.r.t. distance or/and time? Think about what forces are acting on the block, and when. How much energy is required to break the static friction? (Warning, kinetic energy is not conserved in this problem, which is a general feature of friction problems and inelastic collision problems!)

 

[songoku]

Hi turin

You have two basic "errors". (At least, you are not being specific enough for me to tell whether or not you have the correct idea.)

1) What is the final speed of the block? (i.e., does it lose all of its kinetic energy?)

I think it's zero because there is friction

2) What will you consider as the initial kinetic energy of the block? Or, better yet, can you plot the kinetic energy w.r.t. distance or/and time? Think about what forces are acting on the block, and when. How much energy is required to break the static friction? (Warning, kinetic energy is not conserved in this problem, which is a general feature of friction problems and inelastic collision problems!)

Initial kinetic energy of the ball = 0.5*m*v² , where v is the speed after the impact.

The plot of Ek vs distance is straight line with negative slope , and the plot of Ek vs time is a curve that decreases exponentially ??

After the impact, the force that acting on the block is friction (neglecting the vertical forces).

I don't know how to find the energy required to break the friction. We need external force at least the same amount as the friction, but to calculate energy we need distance ??

Thanks