# maple ta quiz 7.3

## 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 21-23)? 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 head-on 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 head-on 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 head-on 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 21-23)? 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 head-on 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 head-on 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 head-on 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

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 = (m1-m2)/ 2m1

then

(2m1)(.9865) = m1-m2

1.973m1 = m1-m2

m2 + 1.973m1 = m1

m2= m1-1.973m1 There is an implied 1 in front of m1 so:

m2 = -.973m1

(m2/m1)= .973

**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 = (m1-m2)/ 2m1

then

(2m1)(.9865) = m1-m2

1.973m1 = m1-m2

m2 + 1.973m1 = m1

m2= m1-1.973m1 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...

(m1-m2)/(m1+m2)

m1 is 1

m2 is the ratio given

(m1-m2)/(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 : 2008-09-19

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