The x components of the average accelerations?

[emily081715]

Homework Statement

A(n) 8.5-kg object is sliding across the ice at 2.34 m/s in the positive x direction. An internal explosion occurs, splitting the object into two equal chunks and adding 12 J of kinetic energy to system. The explosive separation takes place over a 0.16-s time interval. Assume that the one of the chunks after explosion moves in the positive x direction. The x component of the average acceleration of this chunk during the explosion is afrontx, the x component fo the average acceleration of the other chunk during the explosion is arearx. What are the x components of the average accelerations of the two chunks during the explosion?

Homework Equations

k=1/2mv^2

The Attempt at a Solution

k=1/2(8.5)(2.34)^2
= 23.27J

23.27+12=35.27J
35.27/0.16= 220.4m/s^2
i've never done a problem like this

 

[beamie564]

What values are required to compute the average acceleration? ( or what's the formula for it).

 

[emily081715]

What values are required to compute the average acceleration? ( or what's the formula for it).

a=vf-vi/t

 

[beamie564]

a=vf-vi/t

Right. You've been given $t$ and the initial velocity for both the pieces. You need to find their final velocities. There are two unknowns, so we'll need 2 equations to solve them. Can you see what the equations can be?

 

[emily081715]

is one going to be the change in kinetic energy formula, and is the change momentum??

 

[emily081715]

is one going to be the change in kinetic energy formula, and is the change momentum??

is the initial velocity for both pieces just 2.34?

 

[beamie564]

is one going to be the change in kinetic energy formula, and is the change momentum??

Yes. You'll solve them to get the final velocities of both the pieces.

is the initial velocity for both pieces just 2.34?

That's right.

 

[emily081715]

Yes. You'll solve them to get the final velocities of both the pieces.

That's right.

for the mass do i use 8.5, or need to divide that by two since its split equally?

 

[beamie564]

for the mass do i use 8.5, or need to divide that by two since its split equally?

When expressing the final momentum ( or kinetic energy) of the system, you'll have to divide it by two.

 

[emily081715]

When expressing the final momentum ( or kinetic energy) of the system, you'll have to divide it by two.

but on the initial side, it would just be 8.5(2.34) for momentume, and 1/2(8.5)(2.34)^2 for kinetic energy?

 

[beamie564]

but on the initial side, it would just be 8.5(2.34) for momentume, and 1/2(8.5)(2.34)^2 for kinetic energy?

Yes, that's correct.

 

[emily081715]

Yes, that's correct.

i got answers of 20.48 and 25.16. they're wrong

 

[emily081715]

 

[emily081715]

that was my work

 

[beamie564]

Your application of momentum conservation doesn't look right.
Suppose $v$ to be the initial velocity of the block and $v_1$ and $v_2$ be the final velocities of the two pieces. If the mass of the block is $m$, the conservation principle gives $$p_i = p_f $$
$$ mv = \frac {m}{2}v_1 + \frac{m}{2}v_2 $$
This is your first equation. Similarly, for kinetic energy you'll get $$ \frac{1}{2}mv^2 + 12 = \frac{1}{2}\frac{m}{2}(v_1)^2 +\frac{1}{2}\frac{m}{2}(v_2)^2 $$
Solve these two equations for $v_1$and $v_2$. Use them in the formula for average acceleration.

 

[emily081715]

Your application of momentum conservation doesn't look right.
Suppose $v$ to be the initial velocity of the block and $v_1$ and $v_2$ be the final velocities of the two pieces. If the mass of the block is $m$, the conservation principle gives $$p_i = p_f $$
$$ mv = \frac {m}{2}v_1 + \frac{m}{2}v_2 $$
This is your first equation. Similarly, for kinetic energy you'll get $$ \frac{1}{2}mv^2 + 12 = \frac{1}{2}\frac{m}{2}(v_1)^2 +\frac{1}{2}\frac{m}{2}(v_2)^2 $$
Solve these two equations for $v_1$and $v_2$. Use them in the formula for average acceleration.

i got answers of 74.56 and -74.31 for acceleration which are wrong

 

[beamie564]

i got answers of 74.56 and -74.31 for acceleration which are wrong

That doesn't match with my results either.
What values did you get for $v_1$ and $v_2$ ? Show your work.

 

[emily081715]

i go values of 14.27 m/s and -9.59 m/s

 

[beamie564]

In the third line, you have substituted for $v_2$ in terms of $v_1$. You seem to have made a mistake there.
From the first equation(momentum conservation), we get $2v = v_1 + v_2$. Or $v_2 = 4.68 - v_1$ . Use this in your second equation.

 

[emily081715]

i got accelerations of -117.136 m/s^2 and 117.136m/s^2

 

[emily081715]

i got accelerations of -117.136 m/s^2 and 117.136m/s^2

is this correct?

 

[beamie564]

is this correct?

I don't think so. You're one step closer though. The average accelerations you get must have the same magnitude and opposite signs. Show your work for the above result.

 

[emily081715]

 

[emily081715]

what did i miss?

 

[beamie564]

what did i miss?

In the third line you've written $(4.68-v_1)^2$ as $( 21.9014-v^2_1)$. Are you allowed to do that?

 

[emily081715]

In the third line you've written $(4.68-v_1)^2$ as $( 21.9014-v^2_1)$. Are you allowed to do that?

Wouldn't you apply the ^2 to both numbers?

 

[emily081715]

Wouldn't you apply the ^2 to both numbers?

Or would you have to turn it into 21.904-9.36v1 -v1^2

 

[beamie564]

The expansion for $(a-b)^2$ is $(a^2+b^2-2ab)$

 

[emily081715]

That's what this is 21.904-9.36v1-v1^2

 

[emily081715]

i got accelerations now of 7.1 and -7.1 m/s^2
is this correct