Overview

-Complete the following Carolina Biological Lab: Momentum.

Please read the Distance Learning Laboratory Safety Agreement. By participating in this taking part in these labs, you agree to the terms in the Distance Learning Laboratory Safety Agreement. Note: You do not need to submit a signed version of the agreement.

Momentum

This lab explores the behavior of objects in elastic collisions and how differences in mass may affect those collisions.

• Preparation (15 minutes)
• Activity 1: Elastic Collision with Equal Masses (25 minutes)
• Activity 2: Elastic Collision: Mass Added to Cart A (25 minutes)
• Activity 3: Elastic Collision: Mass Added to Cart B (25 minutes)

Note: Additional time (one to two hours) will be needed to complete the lab report following the activities. Read through the Momentum Investigation Manual, including the overview, objectives, and background information.

Review what materials will be needed (p. 6), the Safety procedures (p. 7 ), and acquire the needed materials that are not already supplied.

• -Prepare for the activity by reading through the entire set of instructions prior to starting. The instructions are found on pages 8-18.
• -Set out all materials necessary for this activity that are listed in the Materials section (p. 6) in preparation to perform the activity.
• -Ensure that you have set aside the necessary time to perform the activity. This is listed in the Time Requirements section (p. 3).
• -Perform the activities (pp. 10-18), following the instructions carefully.
• Allow additional time (one to two hours) to complete the lab report, and upload the lab report document when finished.
• Use the PHY 150 M7 Momentum Lab Reporttemplate to complete the lab assignment.

Supporting Materials

This How to - Submit an Assignment tutorial from SNHU Service Now may be helpful if you need guidance when uploading your document.

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Momentum

Student Name

Date

Activity 1: Elastic Collision with Equal Masses Data Table 1

Table 1A. Cart A before collision.

Cart A mass, m (kg)	Distance, d (m)	Time, t (s)	Average time, t (s)	Velocity = d/t (m/s) vA
67.5 g = 0.0675 kg	50 cm = 0.5 m	Trial 1: 0.41	0.456 s	1.09 m/s
Trial 2: 0.47
Trial 3: 0.49

Table 1B. Cart A after collision.

Cart A mass, m (kg)	Distance, d (m)	Time, t (s)	Average time, t (s)	Velocity = d/t (m/s) vA'
67.5 g = 0.0675 kg	0 m	Trial 1: 0	0 s	0 m/s
Trial 2: 0
Trial 3: 0

Table 1C. Cart B after collision.

Cart B mass, m (kg)	Distance, d (m)	Time, t (s)	Average time, t (s)	Velocity = d/t (m/s) vB'
66.1 g = 0.0661 kg	50 cm = 0.5 m	Trial 1: 0.30	0.32 s	1.56 m/s
Trial 2: 0.32
Trial 3: 0.34

Calculations for Activity 1. Elastic Collision with Equal Masses

Apply the law of conservation of momentum to the two-cart system by calculating the momentum before and after the collision.

Helpful equations:

Momentum before the collision = +

Momentum after the collision = +

+ = +

1. Calculate the momentum of the system before the collision (the left side of the equation) and after the collision (the right side of the equation).
2. mA = 0.0675 kg, vA = 1.09 m/s, mB = 0.0661 kg, vB = ?
3. we have given equation.

+ = +

Or 0.0675 kg * 1.09 m/s + 0.0661 kg * vB

Or vB = (0.0675 kg * 1.09 m/s)/0.0661 kg

So, vB = 1.1130 m/s

We know momentum before the collision = mAvA+mBvB

= 0.0675 kg * 1.09 m/s + 0.0661 kg * vB

Substituting the value of vB

0.0675 kg * 1.09 m/s + 0.0661 kg * 1.1130 m/s

= 0.1471 kg m/s

So, Momentum before the collision = 0.1471 kg m/s

Momentum after the collision = +

mA = 0.0675 kg, vA' = 0 m/s, mB = 0.0661 kg, vB' = 1.56 m/s

+ = 0.0675 kg * 0 + 0.0661 kg * 1.56 m/s

=0 + 0.103116 kg m/s

So, Momentum after the collision = 0.103116 kg m/s

1. Calculate the percent difference between the two values.
2. percent difference = _{* 100%={(0.14171 kg m/s - 0.103116 kg m/s)/ ((0.14171 kg m/s +0.103116 kg m/s)/2)} *100%= (0.043984/0.125108) *100%=0.3515 * 100= 35.15%}
3. Explain any difference in the values before and after the collision.

Activity 2: Elastic Collision: Mass Added to Cart A Data Table 2

Table 2A. Cart A before collision.

Cart A mass, m (kg)	Distance, d (m)	Time, t (s)	Average time, t (s)	Velocity = d/t (m/s) vA
204.2 g = 0.2042 kg	50 cm = 0.5 m	Trial 1: 0.43	0.46 s	1.08 m/s
Trial 2: 0.48
Trial 3: 0.47

Table 2B. Cart A after collision.

Cart A mass, m (kg)	Distance, d (m)	Time, t (s)	Average time, t (s)	Velocity = d/t (m/s) vA'
204.2 g = 0.2042 kg	20 cm = 0.2 m	Trial 1: 0.21	0.23 s	0.86 m/s
Trial 2: 0.23
Trial 3: 0.25

Table 2C. Cart B after collision.

Cart B mass, m (kg)	Distance, d (m)	Time, t (s)	Average time, t (s)	Velocity = d/t (m/s) vB'
66.1 g = 0.0661 kg	50 cm 0.5 m	Trial 1: 0.48	0.44 s	1.13 m/s
Trial 2: 0.44
Trial 3: 0.40

Calculations for Activity 2. Elastic Collision: Mass Added to Cart A.

Apply the law of conservation of momentum to the two-cart system by calculating the momentum before and after the collision.

Helpful equations:

Momentum before the collision = +

Momentum after the collision = +

+ = +

1. Calculate the momentum of the system before the collision (the left side of the equation) and after the collision (the right side of the equation).
2. mA = 0.2042 kg, vA = 1.08 m/s, mB = 0.0661 kg, vB = ?, vA' = 0.86 m/s, vB' = 1.13 m/s
3. we use the given equation to find out vB
4. mAvA + mBvB = mAvA' + mBvB'
5. Or 0.2042 kg * 1.08 m/s + 0.0661 kg *vB = 0.2042 kg * 0.86 m/s + 0.0661 kg * 1.13 m/s
6. Or vB = {(0.175612 mg m/s + 0.074693 kg m/s) - 0.220536 kg m/s}/0.0661 kg
7. Or vB = 0.45036 m/s

We have given equationMomentum before the collision = +

=0.2042 kg * 1.08 m/s + 0.0661 kg *vB

Putting the value of vB on the above equation

0.2042 kg * 1.08 m/s + 0.0661 kg * 0.45036 m/s

= 0.220536 kg m/s + 0.029768 kg m/s

= 0.250304 kg m/s

So, momentum before the collision = 0.250304 kg m/s

We have another equation to find out momentum after the collision is

mAvA' + mBvB'

=0.2042 kg * 0.86 m/s + 0.0661 kg * 1.13 m/s

= 0.175612 mg m/s + 0.074693 kg m/s)

= 0.250305 kg m/s

So, Momentum after the collision = 0.250305 m/s

1. Calculate the percent difference between the two values.= = 3.99%
2. Explain any difference in the values before and after the collision.

Activity 3: Elastic Collision: Mass Added to Cart B Data Table 3

Table 3A. Cart A before collision.

Cart A mass, m (kg)	Distance, d (m)	Time, t (s)	Average time, t (s)	Velocity = d/t (m/s) vA
99.0 g = 0.099 kg	50 cm = 0.5 m	Trial 1: 0.47	0.47 s	1.06 m/s
Trial 2: 0.48
Trial 3: 0.46

Table 3B. Cart A after collision.

Cart A mass, m (kg)	Distance, d (m)	Time, t (s)	Average time, t (s)	Velocity = d/t (m/s) vA'
99.0 g = 0.099 kg	-30 cm = 0.3 m	Trial 1: 0.29	0.30 s	1.0 m/s
Trial 2: 0.31
Trial 3: 0.30

Table 3C. Cart B after collision.

Cart B mass, m (kg)	Distance, d (m)	Time, t (s)	Average time, t (s)	Velocity = d/t (m/s) vB'
356.1 g = 0.3561 kg	10 cm 0.1 m	Trial 1: 0.15	0.156 s	0.64 m/s
Trial 2: 0.13
Trial 3: 0.19

Calculations for Activity 3. Elastic Collision: Mass Added to Cart B.

Apply the law of conservation of momentum to the two-cart system by calculating the momentum before and after the collision.

Helpful equations:

Momentum before the collision = +

Momentum after the collision = +

+ = +

1. Calculate the momentum of the system before the collision (the left side of the equation) and after the collision (the right side of the equation).

We can use the given equation + = +

1. To find out vB
2. Given:
3. mA = 0.099 kg, vA = 1.06 m/s, mB = 0.3561 kg, vA' = 1.0 m/s, vB' = 0.64 m/s
4. + = +
5. Or 0.099 kg * 1.06 m/s +0.3561 kg * vB = 0.099 kg * 1.0 m/s +0.3561 kg *0.64 m/s

Or vB = (0.326904 kg m/s - 0.10494 kg m/s)/0.3561 kg

= 0.032211 m/s

So, vB = 0.032211 m/s

Putting the value of vB we can find out momentum before the collision and after the collision.

First, let's find out momentum before the collision

We have given equation:

Momentum before the collision = +

=0.099 kg * 1.06 m/s +0.3561 kg * vB

Putting the value of vB

=0.099 kg * 1.06 m/s +0.3561 kg * 0.032211 m/s

= 0.10494 kg m/s + 0.011470 kg m/s

= 0.116410 kg m/s

So,Momentum before the collision = 0.116410 kg m/s

Let's find out momentum after collision by using the equation +

+

= 0.099 kg * 1.06 m/s + 0.3561 kg m/s * 0.64 m/s

= 0.10494 kg m/s + 0.227904 kg m/s

=0.332844 kg m/s

So, Momentum after the collision = 0.332844 kg m/s

1. Calculate the percent difference between the two values.We have given equation to find out the percent difference as:

= 96.35%

1. Explain any difference in the values before and after the collision.

Questions for Momentum:

1. The law of conservation of momentum states that the total momentum before a collision equals the total momentum after a collision provided there are no outside forces acting on the objects in the system. What outside forces are acting on the present system that could affect the results of the experiments?

2. What did you observe when Cart A containing added mass collided with Cart B containing no mass? How does the law of conservation of momentum explain this collision?

3. In one of the experiments, Cart A may reverse direction after the collision. How is this accounted for in your calculations?