# Length covered by block after collision

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1. Sep 8, 2016

### decentfellow

1. The problem statement, all variables and given/known data
Board $A$ is placed on board $B$ as shown. Both boards slide, without moving w.r.t each other, along a frictionless horizontal surface at a speed of $6 \text{m/s}$. Board $B$ hits a resulting board $C$, "head-on". After the collision, board $B$ and $C$ stick together and the board $A$ slides on top of board $C$ and stops its motion relative to board $C$ in the position shown in the diagram. What is the length (in m) of each board? All three boards have the same mass, size and shape. The coefficient of kinetic friction between boards $A$ & $B$ and b/w $A$ & $B$ is $0.3$.

2. Relevant equations
Impulse-momentum theorem:- $\displaystyle\int{F_{net}dt}=\int{dv}$
$\vec{F}=m\vec{a}$
$f_{max}=\mu N \qquad\qquad\qquad\qquad \text{where, N is the normal reaction acting on the body}$

3. The attempt at a solution

Applying the Impulse momentum theorem on all the blocks we get

$$-\int_{0}^{t}{\mu_km_Agdt}=m_A\int_{6}^{v}{dv}\implies \mu_km_Agt=m_A(6-v)\tag{1}$$
$$-\int_{0}^{t}{(N_3-\mu_km_Ag)dt}=m_A\int_{6}^{v}{dv}\implies (N_3-\mu_km_Ag)t=m_B(6-v)\tag{2}$$
$$\int_{0}^{t}{(N_3+\mu_km_Ag)dt}=m_A\int_{0}^{v}{dv}\implies (N_3+\mu_km_Ag)t=m_Cv\tag{3}$$

From $(1), (2)$ and $(3)$, we get
$$\int_{0}^{t}{N_3dt}=2\int_{0}^{t}{\mu_km_Agdt}\implies N_3=2\mu_km_Ag$$
$$v=\dfrac{9}{2}\text{m/s}$$
$$t=\dfrac{1}{2}\text{sec}$$
$$a_{A/C}=4\mu_kg$$
$${v_{A/C}}_i=6\text{m/s} \qquad\&\qquad {v_{A/C}}_f=0$$
Now, as per the equation of motion for constant acceleration, we have
$$S_{A/C}=\dfrac{-u^2_{A/C}}{2a_{A/C}}=\dfrac{6^2}{8\mu_kg}=\dfrac{6^2}{4\times 3\times 2}=\dfrac{3}{2}\text{m}$$

But, the answer given in the book is $10m$, where am I going wrong.

2. Sep 9, 2016

### Andrew Mason

What is the final velocity of the three boards?

AM

3. Sep 9, 2016

### decentfellow

It is $v=\dfrac{9}{2}\text{m/s}$

4. Sep 9, 2016

### haruspex

I was not able to follow the reasoning by which you arrived at that. Try a simpler way:
- what is the initial total momentum?
- what is the final total momentum?

5. Sep 9, 2016

### decentfellow

Firstly, I had attempted the question the way you told to but at that time I had thought some really weird thing which led me to think that external force was acting on the system after thinking on it again there was no external force acting on it. So, we do get $v=\dfrac{2mv_i}{3m}=\dfrac{2\times6}{3}=4\text{m/s}$, the time $t=\dfrac{2}{3}\text{sec}$, the acceleration of the block C as $a_c=\dfrac{4}{(2/3)}=6\text{m/s}^2$. So, we get $a_{A/C}=-9\text{m/s}^2$, therefore, $S_{A/C}=\dfrac{-u^2_{A/C}}{2a_{A/C}}=\dfrac{6^2}{2\times9}=2\text{m}$

Still, the answer that I get is not the same as the book's.

6. Sep 9, 2016

### BvU

You make it look as if block A moves with the same speed as block B and C right after the collision of B and C ?

7. Sep 9, 2016

### BvU

I don't see a diagram showing this. Is A half way C when no longer sliding ?

8. Sep 9, 2016

### decentfellow

Oh no! I am very sorry I will be uploading the diagram in a moment.

9. Sep 9, 2016

### decentfellow

Here is the diagram accompanying the question

10. Sep 9, 2016

### BvU

OK, so it slides the full length of C. I agree with the 4 m/s for the 'after' case. Check out the situation right after B collides with C.

11. Sep 9, 2016

### decentfellow

what I think is that an impulsive normal reaction force acts on both the blocks $B \& C$ and block $A$ due to inertia continues to move with the same speed it was moving before just that due to relative motion b/w the contact surface a retarding force starts to act on it. I thought that to remain on the safe side I should use the impulse momentum theorem to analyse the situation, but I don't know why but it gives all the wrong values of the variables.

12. Sep 9, 2016

### haruspex

That all seems right. I didn't understand how you got the time 2/3s in post #5.
What are the speeds of A, B and C just after collision? What is the speed of A relative to B&C then? What is the relative acceleration?

13. Sep 9, 2016

### decentfellow

From the
From the fact that a constant force $\mu_km_Ag$ on the block $A$ to finally stop its relative motion with block $C$. So, we get
$$\mu_km_Agt=m_A(6-v)\implies t=\dfrac{6-4}{\mu_kg}=\dfrac{2}{3}$$. Am, I right ?

Just after the collision the speed of the block $A$ should be $6\text{m/s}$ because there is no impulsive normal reaction acting on its surface but there is force of friction acting on it to stop its relative motion but just after the collision, i.e. in a time $dt$, it does not deaccelerate to any significant time.

And for the blocks $B$ & $C$ as there is an external force, i.e. the force of friction, so the momentum of the system is not conserved exactly but due to the same reasoning as that in block $A$, the deaccelration produced due to the friction force is not significant, so we assume the momentum of the system can be assumed to be approximately conserved. So, we get
$$mv=2mv'\implies v'=\dfrac{v}{2}=3\text{m/s}$$

For the relative acceleration what I did was as the block $A$ has a deacceleration of magnitude $\mu_kg$ and from the equation of motion we get $$v_f-v_i=a_Ct\implies 4-3=a_C(2/3)\implies a_C=\dfrac{3}{2}\text{m/s}^2$$ ,so we get $a_{A/C}=4.5\text{m/s}^2$.

So we get $$S_{A/C}=\dfrac{-u^2_{A/C}}{2a_{A/C}}=\dfrac{(6-3)^2}{2\times 4.5}=\dfrac{9}{9}=1m$$

This again brought a new answer, where did I go wrong in this.

Last edited: Sep 9, 2016
14. Sep 9, 2016

### BvU

If this is after the connection of B to C, then what are the 6 and the $v$ ?
Is considering things in the reference frame of C equivalent to working in an inertial frame of reference ?

15. Sep 9, 2016

### decentfellow

$6\text{m/s}$ is the velocity of the blocks $A$ & $B$ before collision and the velocity of the block $A$ just after the collison, and $v$ is the velocity of the system comnprising of $A,B$ and $C$ after the block $A$ has stopped in the position shown in the figure accompanying the question. Also, working in the reference frame of C is not inertial as its an accelerated system so, if analysing the motion of block $A$ we would have to include a pseudo force but I did not do that because I solved everything in the ground's frame of reference and when I had to find the relative displacement I just found out the relative velocity and acceleration.

16. Sep 9, 2016

### J Hann

Why isn't the change in kinetic energy of block A just equal to the work done by friction when A slides a distance L?

17. Sep 9, 2016

### haruspex

Good question. Let's see where that takes us:
Loss of KE of A is m(62-42)/2=10m. Force=3m. L=10/3.
We can do the same for B&C: 2m(42-32)/2=7m. L=7/3.
Interestingly, the difference is 1m, as correctly found to be the answer in post #13.

18. Sep 10, 2016

### Andrew Mason

Total momentum is conserved at ALL times. So immediately after the collision, between B and C, A is still moving at 6 m/sec. B and C travel at 3 m/sec.

We know that the kinetic energy lost to friction is lost as heat. So try working out how much energy is lost between the moment of impact between B and C and the moment that A stops moving relative to B and C. You have to take into account the loss of kinetic energy of A and the gain in kinetic energy of B + C. The difference is the energy lost as friction. Divide that by the force of friction to get the sliding distance (L). Don't assume the book is correct.

AM

19. Sep 10, 2016

### haruspex

Yes, that method is also valid, and produces the same answer decentfellow got in post #13.

20. Sep 10, 2016

### decentfellow

I think its safe to conclude that the answer the book has written is wrong and the answer should be $1\text{m}$ as the work energy theorem also gives the same answer.