maple ta quiz 7.3
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maple ta quiz 7.3
Question 1: (1 point)
A 44.01 gram bullet with a speed of 207.9 m/s hits the 1.001 kg suspended block of a ballistic pendulum and stays embedded in the block. How high does the struck block rise (see sheet 2123)? Indicate with a negative (positive) sign whether the collision between the bullet and the block is perfectly inelastic (elastic).
Question 2: (1 point)
A billiard ball with a velocity of 19.96 m/s makes an elastic headon collision with another ball (same mass) which is initially at rest. Calculate the velocity of the struck ball after the collision (see sheet 25,29). Indicate with a negative (positive) sign whether the incident ball after the collision is (is not) at rest.
question 3: (1 point)
An object makes an elastic headon collision with another "target" object which is initially at rest. If the ratio of incident mass over target mass is 0.8518 what is the velocity of the incident object after the collision in multiples of its incident velocity (see sheet 25,29)? Give the result with the appropriate sign taking the incident velocity as positive.
Question 4: (1 point)
In the experimental discovery of the neutron (the electrically neutral "brother" of the proton) the ratio of the velocity of the recoiling (struck) proton over the velocity of the incident neutron was 1.0135 . What was the ratio of the neutron mass over the proton mass (see sheet 25,29)? (Assume an elastic headon collision).
A 44.01 gram bullet with a speed of 207.9 m/s hits the 1.001 kg suspended block of a ballistic pendulum and stays embedded in the block. How high does the struck block rise (see sheet 2123)? Indicate with a negative (positive) sign whether the collision between the bullet and the block is perfectly inelastic (elastic).
Question 2: (1 point)
A billiard ball with a velocity of 19.96 m/s makes an elastic headon collision with another ball (same mass) which is initially at rest. Calculate the velocity of the struck ball after the collision (see sheet 25,29). Indicate with a negative (positive) sign whether the incident ball after the collision is (is not) at rest.
question 3: (1 point)
An object makes an elastic headon collision with another "target" object which is initially at rest. If the ratio of incident mass over target mass is 0.8518 what is the velocity of the incident object after the collision in multiples of its incident velocity (see sheet 25,29)? Give the result with the appropriate sign taking the incident velocity as positive.
Question 4: (1 point)
In the experimental discovery of the neutron (the electrically neutral "brother" of the proton) the ratio of the velocity of the recoiling (struck) proton over the velocity of the incident neutron was 1.0135 . What was the ratio of the neutron mass over the proton mass (see sheet 25,29)? (Assume an elastic headon collision).
guest121 Guest
Question 1
For Question one use the formula
v = sqrt(2gh)(M+m/m) and rearrange the formula to get h by itself.
When you are ready, the formula should look like this:
(mV/M+m)^2/2g = h  make sure you do the parenthesis and square it first THEN divide by 2g
My equation was:
m = 26.65g = .02665kg
M = 1.394kg
g = 9.81 m/s^2
v = 237.1 m/s
((.02665*237.1)/(.02665+1.394))^2/2g
(6.223/1.420)^2 = 19.19 ...... 2(9.81) = 19.62
19.19/19.62 = .9785
ANSWER IS NEGATIVE
v = sqrt(2gh)(M+m/m) and rearrange the formula to get h by itself.
When you are ready, the formula should look like this:
(mV/M+m)^2/2g = h  make sure you do the parenthesis and square it first THEN divide by 2g
My equation was:
m = 26.65g = .02665kg
M = 1.394kg
g = 9.81 m/s^2
v = 237.1 m/s
((.02665*237.1)/(.02665+1.394))^2/2g
(6.223/1.420)^2 = 19.19 ...... 2(9.81) = 19.62
19.19/19.62 = .9785
ANSWER IS NEGATIVE
Guest 00 Guest
Question 3
Here is how to solve question 3. I am going to use my own values but you should be able to plug in your values.
Question 3
You need to use both momentum conservation and elastic condition.
Okay my ratio was .5002 and the given correct answer was .3331555792
(1) mv_1 = mv_2 + mv_3 momentum conservation
(2) .5mv_1^2 = .5mv_2^2 + .5mv_3^2 elastic condition
then
(1) .5002mv_1 = .5002mv_2 + mv_3
(2) 5(.5002m)v_1^2 = .5(.5002mv_2^2 + mv_3^2)
Then cancel out common values
(1) .5002 = .5002v_2/v_1 + v_3/v_1
(2) .5002 = .5002v_2^2/v_1^2 + v_3^2/v_1^2
x=v_2/v_1
y=v_3/v_1
(1) .5002 = .5002x + y
(2) .5002 = .5002x^2 +y^2
Solve for y in (1)
y = .5002  .5002x
plug into (2)
.5002 = .5002x^2 + (.5002.5002x)^2
.5002 = .5002x^2 + .25020004 +.25020004x^2
.24999996 = .5002x^2 +.25020004x^2
.24999996 = .75040004x^2
x^2 = .333155579
This is a negative value
Question 3
You need to use both momentum conservation and elastic condition.
Okay my ratio was .5002 and the given correct answer was .3331555792
(1) mv_1 = mv_2 + mv_3 momentum conservation
(2) .5mv_1^2 = .5mv_2^2 + .5mv_3^2 elastic condition
then
(1) .5002mv_1 = .5002mv_2 + mv_3
(2) 5(.5002m)v_1^2 = .5(.5002mv_2^2 + mv_3^2)
Then cancel out common values
(1) .5002 = .5002v_2/v_1 + v_3/v_1
(2) .5002 = .5002v_2^2/v_1^2 + v_3^2/v_1^2
x=v_2/v_1
y=v_3/v_1
(1) .5002 = .5002x + y
(2) .5002 = .5002x^2 +y^2
Solve for y in (1)
y = .5002  .5002x
plug into (2)
.5002 = .5002x^2 + (.5002.5002x)^2
.5002 = .5002x^2 + .25020004 +.25020004x^2
.24999996 = .5002x^2 +.25020004x^2
.24999996 = .75040004x^2
x^2 = .333155579
This is a negative value
Guest 11 Guest
Question 4
Here is how to solve question 4 The key to this question is that it say that this is an elastic head on collision.
We already have the tools to solve this question. Since this is a head on collision this is in one Dimension.
You have to use the equation from sheet 27 and sheet 28. The struck proton is v1 and the neutron is v2. This also means that the neutron mass is m2 and the proton mass is m1. Since we are given the ratio (.9865 in my case) as being the final velocity of the proton over the neutron, we can divide the equation from sheet 27 over the equation from sheet 28. The given correct answer in my case was .973
I'm going to skip the actual division since that is fairly simple but the v1 from both equations cancel out and you should get something like:
.9865 = (m1m2)/ 2m1
then
(2m1)(.9865) = m1m2
1.973m1 = m1m2
m2 + 1.973m1 = m1
m2= m11.973m1 There is an implied 1 in front of m1 so:
m2 = .973m1
(m2/m1)= .973
We already have the tools to solve this question. Since this is a head on collision this is in one Dimension.
You have to use the equation from sheet 27 and sheet 28. The struck proton is v1 and the neutron is v2. This also means that the neutron mass is m2 and the proton mass is m1. Since we are given the ratio (.9865 in my case) as being the final velocity of the proton over the neutron, we can divide the equation from sheet 27 over the equation from sheet 28. The given correct answer in my case was .973
I'm going to skip the actual division since that is fairly simple but the v1 from both equations cancel out and you should get something like:
.9865 = (m1m2)/ 2m1
then
(2m1)(.9865) = m1m2
1.973m1 = m1m2
m2 + 1.973m1 = m1
m2= m11.973m1 There is an implied 1 in front of m1 so:
m2 = .973m1
(m2/m1)= .973
Guest 11 Guest
Question 3
For number 3, use this instead...
(m1m2)/(m1+m2)
m1 is 1
m2 is the ratio given
(m1m2)/(m1+m2)
m1 is 1
m2 is the ratio given
Guest00 Guest
question 2
Does anyone know how to solve number two?
I've tried but I can't seem to get it....
Thank you!
I've tried but I can't seem to get it....
Thank you!
stuckk Guest
Re: maple ta quiz 7.3
For Question 2 the velocities are equal. S the ball traveling at 120mph hits the other ball in a perfectly elastic collision and the second ball continues at the same speed.
The answer is negative as well.
The answer is negative as well.
Guest01 Posts : 133
Join date : 20080919
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