Momentum

Experiment 9 Conservation of Linear Momentum - Collisions In this experiment you will be introduced to the definition of linear momentum. You will learn the difference between an elastic and an inelastic collision. You will explore how to determine the amount of momentum before and after an elastic and an inelastic collision. In order to study these effects you will use a motion sensor. For the collision you will use the frictionless track and carts equipped with the necessary attachments to undergo elastic and inelastic collisions. Objectives: 1. Measure the velocities of the carts before and after the collisions. 2. Calculate the momentum and the kinetic energies of the cart before and after the collisions. 3. In the case of inelastic collisions, you will verify the law of conservation of momentum, viz. compare the momentum before and after the collision for each cart based on the measured masses and velocities. 4. In the case of elastic collisions, you will verify the law of conservation of momentum, viz. compare the momentum before and after the collision for each cart based on the measured masses and velocities. You will also study the effect of such a collision on the kinetic energy of each cart. Hypothesis: How does the momentum before an elastic and an inelastic collision compare to the momentum after the collision? If two objects with equal but opposite momenta collide head-on inelastically, what is their shared velocity after the collision? Theory: If a mass m moves along a straight line with a velocity v, the linear momentum of the mass is defined as p = mv. Momentum is a vector quantity and thus has a magnitude and a direction. Inelastic collisions: Figure 1 shows two carts with mass m1 and m2 and with initial velocities v1i and v2i respectively headed for a collision (v1i > v2i) in one dimension (i.e the motions before and after the collision are along a single axis). The two carts form the system under consideration that is isolated from other effects. We write the law of conservation of linear momentum for this two-body system as the following:

1

After

Before v1i

v2i

m1

m2

v1f

v2f

m1

m2

Figure 1

Total momentum before the collision = Total momentum after the collision m1 v 1i + m2 v 2i = m1 v 1 f + m2 v 2 f

(1)

where v1f and v2f are the final velocities of the carts after the collision. If the initial velocities of the carts is known and one final velocity is known then using the equation above we can calculate the other final velocity.

After

Before v1i m1

v2i = 0 m2

V m1

m2

Figure 2

In the case of a completely inelastic collision the two carts stick together after the collision as seen in Figure 2. In this case we can let the second cart (m2) stay at rest (v1i = 0) while the first cart moves with an initial velocity v1i. After the collision the two masses stick together and move with a common velocity V. Conservation of momentum requires that:

m1v1i = ( m1 + m2 )V (2) m1v1i m1 + m2 In the equation above, if we know the masses and the initial velocity, we can calculate the velocity of the masses after the collision. As expected, this velocity will be less than the V =

2

m1 must be less than unity. In the inelastic m1 + m 2 collision studied above, the linear momentum is conserved.

initial velocity of the first mass since

Elastic collisions:

After

Before v1i

v2i

m1

m2

v1f

v2f

m1

m2

Figure 3

Figure 3 shows two carts with mass m1 and m2 and with initial velocities v1i and v2i respectively headed for a collision in one dimension (i.e the motions before and after the collision are along a single axis). The two carts form the system under consideration that is isolated from other effects. We write the law of conservation of linear momentum for this two-body system as the following: m1 v 1i + m2 v 2i = m1 v 1 f + m2 v 2 f

(3)

We write the law of conservation of energy for this two-body system as the following:

1 1 1 1 2 2 2 2 m1v1i + m2 v2i = m1v1 f + m2 v2 f 2 2 2 2

(4)

where v1f and v2f are the final velocities of the carts after the collision. Equations (3) and (4) can be solved simultaneously to give the following two equations for the velocities after the collision in terms of the velocities before the collision. v1 f =

m1 − m2 2m2 v1i + v2i m1 + m2 m1 + m2

(5)

v2 f =

m − m1 2m1 v1i + 2 v 2i m1 + m2 m1 + m2

(6)

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Exercises:

1. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere 1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2 with mass m2 = 75 g. What is the velocity of sphere 1 just before the collision (v1i)?

h1

m1

m2

h2

From conservation of energy we have:

1 2 m1v1i = m 1 gh1 2 v1i = 2 gh 1 = 2 * 9 .8 * 0 .080 = 1 .252 m / s 2. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere 1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2 with mass m2 = 75 g. What is the velocity of sphere 1 just after the collision (v1f)? Use Equation (5) v1 f =

m1 − m2 0.030 − 0.075 v1i = * 1.252 = −0.537m / s m1 + m2 0.030 + 0.075

3. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere 1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2 with mass m2 = 75 g. What is the velocity of sphere 2 just after the collision (v2f)?

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Use Equation (6) v2 f =

2m1 2 * 0.030 v1i = * 1.252 = 0.715m / s m1 + m2 0.030 + 0.075

4. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere 1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2 with mass m2 = 75 g. To what height h1’ does the sphere 1 swing to the left after the collision? Use the conservation of energy here to get: 1 2 m1v1 f 2

'

m1 gh1 = '

h1 =

v1 f

2

=

2g

( −0.537) 2 = 0.0147m = 1.5cm 2 * 9.8

5. Two metal spheres, suspended by vertical cords, initially touch each other. Sphere 1 with mass m1 = 30 g is pulled to the left to a height h1 = 8.0 cm and then released from rest. After swinging down, it undergoes an elastic collision with sphere 2 with mass m2 = 75 g. To what height h2 does the sphere 2 swing to the right after the collision? Use the conservation of energy to get: 1 2 m2 v 2 f 2

m2 gh2 = h2 =

v2 f

2

2g

=

(0.715) 2 = 0.0261m = 2.6cm 2 * 9.8

5

M m

h

v

6. A ballistic pendulum is a device that was used to measure the speeds of bullets before electronic timing devices were invented. The device consists of a large block of wood of mass M = 5.4 kg, hanging from two long cords. A bullet of mass m = 9.5 g is fired into the block, coming quickly to rest. The block+bullet then swing upward, their center of mass rising a vertical distance h = 6.3 cm before the pendulum comes momentarily to rest at the end of its arc. What was the speed v of the bullet before it hit the block (v1i)? Use the conservation of linear momentum: m1v1i = ( m1 + m2 )V v1i =

( m1 + m2 )V m1

where m1 = m = 9.5g and m2 = M = 5.4 kg The next step is to find V, the velocity of the block+bullet just after impact. Since the mechanical energy (kinetic plus potential) is conserved after the collision, we have: 1 ( m + M )V 2 = ( m + M ) gh 2 V = 2 * g * h = 2 * 9.8 * 0.063 = 1.111m / s

Now we can find the speed of the bullet before it hit the block as follows ( m + m2 )V (0.0095 + 5.4) v1i = 1 = * 1.111 = 632.6m / s m1 0.0095

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7. What is the linear momentum of an automobile with weight 1000 kg traveling at 60 km/hr?

p = mv = 1000 * 60 * 1000 / 3600 = 16666.67kg .m / s 8. Suppose that your mass is 80 kg. How fast would you have to run to have the same momentum as a 1600 kg car moving at 1.2 km/hr? m1v1 = m2 v 2 v1 =

5.5 m/s

1600 * 1.2 = 24km / hr 80

2.5 m/s

1.6 kg

V2f=?

2.4 kg

4.9 m/s

1.6 kg

BEFORE

2.4 kg

AFTER

9. Two blocks slide on a frictionless surface and collide. What is the velocity of the 1.6 kg block after the collision? Since momentum is conserved we have: m1v1i + m2 v2i = m1v1 f + m2 v 2 f v1 f = =

m1v1i + m2 v 2i − m2 v2 f m1

1.6 * 5.5 + 2.4 * 2.5 − 2.4 * 4.9 = 1.9m / s 1.6

Now we calculate the kinetic energy of the system before the collision and after the collision as follows:

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Before : After :

1 1 1 1 2 2 m1v1i + m2 v2i = * 1.6 * 5.52 + * 2.4 * 2.52 = 31.7 J 2 2 2 2 1 1 1 1 2 2 m1v1 f + m2 v 2 f = * 1.6 * 1.9 2 + * 2.4 * 4.9 2 = 31.7 J 2 2 2 2

10. Is the collision elastic? Yes, the collision is not elastic as the kinetic energy before and after are the same.

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