Momemtum

Lesson 8-Questions

Lesson Notes

Questions

How do I calculate the velocity using the conservation of linear momentum’s formula if there are three objects involved?

Let us use the following as an example:

A bullet of mass 50 g moves at 40 m/s and strikes a wooden block of mass 0, 5 kg that is mounted on a stationary trolley of mass 2 kg. Calculate the velocity of the bullet-wooden block-trolley system if they move as one unit after the collision.

Mass of the bullet  =  50 g  ÷  1000  =  0, 05 kg

∑p_i  =  ∑p_f

m_Av_iA  +  m_Bv_iB  +  m_cv_ic   =   m_Av_fA  +  m_Bv_fB  +  m_cv_fc

(0, 05)(40)  +  (0,5)(0)  +  (2)(0)  =  (0, 05 + 0,5 + 2)(v_f)

2  +  0  +  0  =  (2,55)v_f

v_f  =  (2)/(2,55)

=  0,78431... m/s

=  0, 78 m/s in the initial direction of motion

(Always round your answers off to two decimal places.)

Additional example on conservation of momentum:

A trolley with a mass of 2 kg is moving at five m/s. A brick of mass 500 g is dropped vertically onto the trolley. Calculate the velocity of the trolley and the brick afterwards. By means of calculations show whether it was an elastic or an inelastic collision.

For the brick:

When an object is dropped it means its initial velocity is zero.

The SI unit for mass in physics is kg. In order to convert g to kg we divide by 1000.

m  =  500g  ÷ 1000  =  0, 5 kg

∑pi  =  ∑pf mAviA  +  mBviB   =   mAvfA  +  mBvfB (2)(5)  +  (0,5)(0)  =  (2 + 0,5)vf 10  +  0  =  (2,5)vf Vf  =  (10)/(2,5) =   4 m/s in the initial direction of motion

Because the brick and the trolley are moving as one unit with the same velocity afterwards, we add their masses together.

Elastic/inelastic collision?

E_k before collision    =   ½ m_Av²_iA  +  ½ m_Bv²_iB

=  ½ (2)(5)²  +  ½ (0,5)(0)²

= 25  +  0

= 25 J

E_k after collision    =   ½ m_Av²_fA  + ½ m_Bv²_fB

=  ½ (2)(4)²  +  ½ (0,5)(4)²

=  16  +  4

=  20 J

E_k before collision   ≠  E_k after collision    →   inelastic collision

25 J   ≠   20 J

EXAM-BASED QUESTIONS

Lesson 1

A delivery van with a mass of 4000 kg moves towards the right at a constant velocity of 15 m/s. The delivery van collides head-on with a car of mass 2000 kg moving at a constant velocity of 20 m/s. Immediately after the collision, the car moves at a constant velocity of 5 m/s to the right.

Calculate the magnitude of the velocity of the delivery van immediately after the collision.

Lesson 2

A car with mass m travelling at a velocity of 10 m/s west and a truck of mass 2m travelling at 10 m/s east on the same road, collide. Ignore the effects of friction. After the collision they move as one unit with a velocity of 3,33 m/s east.

On impact the car exerts a force of magnitude F on the truck and it experiences an acceleration of magnitude a.

1.Determine, in terms of F, the magnitude of the force that the truck exerts on the car on impact. Give a reason for your answer.

2.Determine in terms of a acceleration that the truck experiences on impact. Give a reason.

3.Although both drivers are wearing seatbelts, which driver is likely to be more severely injured?

Lesson 3

A space shuttle, consisting of a rocket motor with a mass of 800 kg and a capsule with a mass of 200 kg, is travelling in space at 7 000 m/s relative to the Earth. It releases its rocket motor. As a result, the capsule is projected in the opposite direction at 7 500 m/s relative to the Earth.

Is the collision elastic or inelastic? Support your answer with a calculation.

Lesson 4

John (mass 70 kg) and Pete (65 kg) are standing on a stationary skateboard with a mass of 20 kg, as indicated in the diagram. John and Pete jump off the skateboard simultaneously. John jumps to the left at two m/s and Pete to the right at 1 m/s. What is the velocity of the skateboard after the jump?

ANSWERS

Lesson 1

∑p_i  =  ∑p_f

m_Av_iA  +  m_Bv_iB   =   m_Av_fA  +  m_Bv_fB

(4000)(15)  +  (2000)(-20)  =  (4000)v_fA+   (2000)(5)

60 000  -  40 000=  (4000) v_fA  +  10 000

10 000  =  (4000) v_fA

v_fA  =  (10000)/(4000)

v_fA  =  2,5 m/s to the right

Lesson 2

1.F - According to Newton’s 3^rd Law: Body A exerts a force on body B. Body B exerts the same force on body A in an opposite direction.

2.½ a - According the Newton’s 2^nd Law: At a constant force the mass is inversely proportional to the acceleration. Thus, when the mass of the truck is double that of the car, the truck’s acceleration is halved.

3.We need to check which one has the greater change in velocity and acceleration.

Truck driver’s change in velocity: Δv  =  vf  -  viΔv  =  vf  -  vi =  (3,33)  -  (10) =  -6,67 m/s

Car driver’s change in velocity =  (3,33) - (-10) =  13,33 m/s

Thus the driver of the car has a greater change in velocity; he is more likely to sustain severe injuries.

His acceleration is also greater because a = Δv/t and the contact time during the collision is the same for both the truck and the car.

Also, according to Newton’s 2^nd Law, the car’s mass is half that of the truck. Thus it experiences double the acceleration.

Lesson 3

In order to complete the calculation, we first need to calculate the velocity of the rocket after it has been released.

∑p_i  =  ∑p_f

m_A+B(v_iA + v_iB)   =   m_Av_fA  +  m_Bv_fB

(800 + 200)(7 000)  =  (800)v_fA+   (200)(-7 500)

7 000 000 =  (800) v_fA  - 1500 000

8 500 000  =  (800) v_fA

v_fA  =  (8 500 000)/(800)

v_fA  =  10 625 m/s in the direction of initial motion

Now we are able to calculate the kinetic energy before and after the collision.

E_k before collision    =   ½ m_Av²_iA  +  ½ m_Bv²_iB

=  ½ (800)(7 000)²  +  ½ (200)(7 000)²

= (1,96 × 10¹⁰)  +  (4,9 × 10⁹)

= 2,45 × 10¹⁰ J

E_k after collision    =   ½ m_Av²_fA  + ½ m_Bv²_fB

=  ½ (800)(10 625)²  +  ½ (200)(-7 500)²

=  (4,52 × 10¹⁰)  +  (5,63 × 10⁹)

=  5,08 × 10¹⁰ J

E_k before collision   ≠  E_k after collision   →      inelastic collision

Lesson 4

∑p_i  =  ∑p_f

m_Av_iA  +  m_Bv_iB  +  m_cv_ic   =   m_Av_fA  +  m_Bv_fB  +  m_cv_fc

(70)(0)  +  (20)(0)  +  (65)(0)  =  (70)(-2) + (20) v_fB + (65)(1)

0  =  (20)v_f - 75

v_f  =  (75)/(20)

=  3, 75 m/s to the right