Momentum and Impulse

Principles of Physics

 - property of an object related to its mass and

velocity.- “mass in motion” or “inertia in motion” p = momentum (vector)p = mv m = mass (kg)

v = velocity (m/s) (vector) units for momentum are kg m/s **If an object is not moving, it has no momentum

Momentum

Momentum of an individual objectp = mv

Example: Calculate the momentum of a 250 kg cart with a velocity of 25 m/s.

p = mvp = 250 kg (25 m/s)p = 6250 kg m/s

Momentum of a systemp = m1v1 + m2v2 + m3v3 + …

Example: A 300 kg car is travelling to east at 45 m/s and a 500 kg truck is travelling west at 30 m/s. Calculate the total momentum of the system.

p = m1v1 + m2v2 p = 300 kg(45 m/s) + 500 kg(-30 m/s)p = -1500 kg m/s

Conservation – to keep constant even though changes occur◦ whatever you had before an event you will still

have after the event

Conservation

The total momentum of a system is the same before and after a collision

ptotal before = ptotal after

Conservation of Momentum

ptotal before = ptotal after

ptotal before= m1v1i + m2v2i + m3v3i + …

 **remember v has direction

 ptotal after = m1v1f + m2v2f + m3v3f + …

Newton’s Cradle Demo:

Conservation of Momentum

Types of Collisions

Elastic - bounce ◦ objects hit and bounce off from each other

Inelastic – stick◦ multiple objects hit and stick together

or◦ one objects separates into 2 or more (explosion)

Example 1A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Givens:

m1 = 0.050 kgm2 = 0.050 kgv1i= 0.20 m/s v2i= 0.10 m/sv1f = 0.08 m/s v2f = ?

Write down given information

Example 1A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Givens:m1 = 0.050 kgm2 = 0.050 kgv1i= 0.20 m/s v2i= 0.10 m/sv1f = 0.08 m/s v2f = ?

Determine the momentum of the cue ball before and after the collision

p1i = m1v1i

= 0.050 kg(0.20 m/s) = 0.01 kg m/s

p1f = m1v1f

= 0.050 kg(0.08 m/s) = 0.004 kg m/s

Example 1A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Givens:m1 = 0.050 kgm2 = 0.050 kgv1i= 0.20 m/s v2i= 0.10 m/sv1f = 0.08 m/s v2f = ?

Determine the momentum of the 8 ball before the collision

p2i = m2v2i

= 0.050 kg(0.10 m/s) = 0.005 kg m/s

Example 1 A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Set up a conservation equation

pbefore = pafter

p1i + p2i = p1f + p2f

Example 1 A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Substitute momentum values into the conservation equation and solve for the momentum of the 8 ball after the collision

pbefore = pafter

p1i + p2i = p1f + p2f

0.01 kg m/s + 0.005 kg m/s = 0.004 kg m/s + p2f p2f = 0.011 kg m/s

Example 1 A cue ball with mass 0.050 kg moves at a velocity of 0.20 m/s. It collides with the 8 ball (0.050 kg) moving in the same direction with velocity 0.10 m/s. After the collision, the cue ball is moving with a velocity of 0.08 m/s in the same direction. What is the 8 ball’s final velocity?

Determine the final velocity of the 8 ball

p2f = m2v2f

0.011 kg m/s = 0.050 kg (v2f) v2f = 0.22 m/s

Example 2A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Givens:mb = 0.015 kgmbb = 5.085 kgvbi= ?vbbi= 0 m/svbf = 1 m/svbbf = 1 m/s

Write down given information

Example 2A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Determine the momentum of the block before and after the collision

p2i = m2v2i

= 5.085 kg(0 m/s) = 0 kg m/s

p2f = m2 v2f

= (5.085 kg) (1 m/s) = 5.085 kg m/s

Givens:m1 = 0.015 kgm2 = 5.085 kgv1i= ?v2i= 0 m/sv1f = 1 m/sv2f = 1 m/s

Example 2A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Determine the momentum of the bullet after the collision

p1f = (m1) v1f

= (0.015 kg) (1 m/s) = 0.015 kg m/s

Givens:m1 = 0.015 kgm2 = 5.085 kgv1i= ?v2i= 0 m/sv1f = 1 m/sv2f = 1 m/s

Example 2A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

pbefore = pafter

p1i + p2i = p1f + p2f

Set up a conservation equation

Example 2A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

pbefore = pafter

p1i + p2i = p1f + p2f

p1i + 0 = 0.015 kg m/s + 5.085 kg m/s p1i = 5.1 kg m/s

Substitute the momentum values and determine the momentum of the bullet before the collision

Example 2A 0.015 kg bullet is shot into a 5.085 kg wood block at rest on a frictionless surface. The block and the bullet acquire a speed of 1 m/s as a result of the collision. Calculate the initial velocity of the bullet.

Determine the initial velocity of the bullet

p1i = m1v1i

5.1 kg m/s = 0.015 kg(v1i) v1i = 340 m/s

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