Conservation of energy word problem

[Alameen Damer]

Homework Statement

A 9.1 g ball is hit into a 98 g block of clay at rest on a level surface. After impact, the block slides 8m before coming to rest. If the coefficient of friction is 0.60, determine the speed of the ball before impact

Homework Equations

Ek+1/2 mv^2
Ff=uFn

The Attempt at a Solution

First of all, I calculate friction force.

Fn=(0.098)(9.8)
=0.96
Ff=(Fn)(Uk)
=5.9 N

I find acceleration of the block:

-5.9 N = 0.098a
-60 m/s^2=a

At this point I am stuck i am unsure of what to find next given that i know no velocities, except the initial velocity for the block. This problem i should solve using conservation of energy and momentum.

 

[haruspex]

As I read it, the ball goes into the clay.
Not sure why you detemined the acceleration, since you had the force and the distance. Anyway, you say you have found the velocity just after impact (though you don't show that). What quantity is conserved during the impact?

 

[Alameen Damer]

Initial velocity as in the velocity of the block at rest, before getting hit. And from your understanding does this seem to be an inelastic collision?

 

[haruspex]

Initial velocity as in the velocity of the block at rest, before getting hit. And from your understanding does this seem to be an inelastic collision?

Ok, so can you find the velocity just after impact?
The collision is certainly inelastic. As I wrote, I believe the question statement implies the ball becomes embedded in the clay. Were this not so, you would not have enough information.

 

[Alameen Damer]

Do we have enough information for this? I attempted to solve it using the conservation of momentum, however I am missing the initial velocity of the ball.

 

[haruspex]

Do we have enough information for this? I attempted to solve it using the conservation of momentum, however I am missing the initial velocity of the ball.

From what you worked out in the OP you can find the velocity immediately after collision. (Your calculation was not really correct because you did not have the right mass for the clay plus ball combination, but, as it happens, this won't change the calculation of the velocity just after collision as long as you are consistent in that regard.)
For the conservation of momentum, the initial velocity of the ball is the only variable that you do not know the value of, so you can deduce it.

 

[Alameen Damer]

Ok so using kinematics can I use the vf^2=vi^2+2ad equation in this case. And if so, would vf be 0 as they both come to rest at the end.

 

[Alameen Damer]

Okay so using vf^2=vi^2+2ad:

Note I found acceleration using the same method as in the OP, just changed the mass to include the ball:

vf^2=vi^2+2ad
0=vi^2+2(-55.1)(8)
881.6=vi^2
29.7=vi

 

[haruspex]

Ok so using kinematics can I use the vf^2=vi^2+2ad equation in this case. And if so, would vi be 0 as the block and ball started from rest, essentially.

No.
You need to split the process into two phases. First, there's the collision. Each object has an initial and final velocity for that phase. The final velocities for that phase are the initial velocities for the second phase, the block plus ball combination sliding to rest.
You found the deceleration for the second phase, and you know the distance and final velocity of that phase, so what is the initial velocity of that phase?

 

[Alameen Damer]

Yes my apologies, I edited my above quote, the Vf for the second phase would be 0, and thus using kinematics as shown above, the initial velocity of the ball+clay during the second phase is 29.7 m/s.

 

[haruspex]

Yes my apologies, I edited my above quote, the Vf for the second phase would be 0, and thus using kinematics as shown above, the initial velocity of the ball+clay during the second phase is 29.7 m/s.

That's much too big. Looks like you made an error in calculating the frictional force.
I strongly recommend you get into the habit of working purely symbolically. Pretend you are not given any numeric values, so you have to create a variable name for each of those. This has many advantages, particularly when asking others to review your work. Only plug in numbers right at the end.

 

[Alameen Damer]

EDITED

This is how I worked it out:

Force of friction:

Note mass of ball+block=0.1071 kg

Ff=(u)(Fn)
=(0.6)(1.04958)
=0.63


Finding acceleration:
-0.63=0.1071a
-5.88=a

Vf^2=Vi^2+2ad
Vi^2=Vf^2-2ad
Vi^2=0-(2)(-5.88)(8)
Vi=9.7 m/s

That is the initial velocity for the second phase, aka final velocity for the first phase. Are the calculations correct?

 

[Alameen Damer]

edited.

 

[haruspex]

EDITED

This is how I worked it out:

Force of friction:

Note mass of ball+block=0.1071 kg

Ff=(u)(Fn)
=(0.6)(1.04958)
=0.63


Finding acceleration:
-0.63=0.1071a
-5.88=a

Vf^2=Vi^2+2ad
Vi^2=Vf^2-2ad
Vi^2=0-(2)(-5.88)(8)
Vi=9.7 m/s

That is the initial velocity for the second phase, aka final velocity for the first phase. Are the calculations correct?

That's better. I guess you had multiplied by g twice. This is an example of the many benefits of working symbolically:
$F_f=\mu F_N=\mu (m_b+m_c)g = a(m_b+m_c)$. So $a=\mu g$. No risk of doubling up the g's, and the masses cancel, reducing the effort.

 

[Alameen Damer]

Yes it is much more organized, and reduces the risk of errors.

At the point of finding this 9.7 m/s velocity, I attempted to sub it into an equation for the final velocity of an inelastic collision-then isolate Vi1:

Inelastic collision:

Vf=[(m1)(vi1)+(m2)(vi2)] / (m1+m2)
Vf(m1+m2)=(m1)(vi1)+(m2)(vi2)

Vi2 is 0 so:

Vf(m1+m2)=(m1)(vi1)
[Vf(m1+m2)]/m1=vi1
[9.7(0.0091+0.0980]/0.0091=vi1
114.2=vi1

The book answer is 33 m/s though

 

[haruspex]

Yes it is much more organized, and reduces the risk of errors.

At the point of finding this 9.7 m/s velocity, I attempted to sub it into an equation for the final velocity of an inelastic collision-then isolate Vi1:

Inelastic collision:

Vf=[(m1)(vi1)+(m2)(vi2)] / (m1+m2)
Vf(m1+m2)=(m1)(vi1)+(m2)(vi2)

Vi2 is 0 so:

Vf(m1+m2)=(m1)(vi1)
[Vf(m1+m2)]/m1=vi1
[9.7(0.0091+0.0980]/0.0091=vi1
114.2=vi1

The book answer is 33 m/s though

Your answer is correct given the numbers you quoted. The 8m seems rather a long way. Are you sure it didn't say 0.8m? But that still gives me 36m/s, not 33.

 

[Alameen Damer]

Yep I'm sure, i have attached the question photo as well. So is my answer correct?

 

[haruspex]

Yep I'm sure, i have attached the question photo as well. So is my answer correct?

Yes. 33m/s is nowhere near enough.

 

[Alameen Damer]

Thank you very much for your help

 

[haruspex]

Thank you very much for your help

You are welcome. For what it's worth, 9.1g is too light for a golf ball. 40-45g would be reasonable. 41g gives 33m/s.