Energy and Momentum Ballistic Pendulum Other Quandary v

Energy and Momentum • Ballistic Pendulum • Other

Quandary v = 560 m/s m = 100. grams bullet Calculate: • Velocity after • Total energy before • Total energy after (1/2 mv 2) Before v = 0 m/s m = 845 grams Block O Wood After v = ? ? ? Bullet and Block combo

So for momentum, the equation looks like (. 1 kg)(560 m/s) = (. 845 kg)v, so v = 59. 25925926 Energy before is ½(. 1 kg)(560 m/s)2 = 15, 680 J Energy after is ½(. 945 kg)(59. 25925926 m/s)2 = 1659. 259259 J (The repeating decimal is creepy) So energy is conserved, just not kinetic energy. Some of the original kinetic energy turns to sound and heat.

Elastic Bouncy Ke is conserved spring bumper Molecules reversible Inelastic Sticky Ke is Not conserved Bullet in wood car bumper v>5 mph irreversible

Before After 351 m/s 0. 012 kg 3. 215 kg Find how high the block will rise after the bullet hits. Hint in the collision momentum is conserved, and when the block and bullet swing up, energy is conserved

So the bullet will impart its momentum to the block: (0. 012 kg)(351 m/s) = (0. 012 kg + 3. 215 kg)(v) v = 1. 305 m/s Now energy will be conserved as it swings like a pendulum: 1/ mv 2 = mgh 2 1/ v 2 = gh 2 1/ (1. 305 m/s)2 = (9. 81 N/kg)h 2 h = 0. 087 m

Before After v=? . 0065 kg 3. 215 kg Δh =. 042 m Find the initial velocity of the bullet before it hits the block. Hint - in the collision momentum is conserved, and when the block and bullet swing up, energy is conserved 450 m/s

To rise. 042 m, the bullet and block must have had this velocity: 1/ mv 2 = mgh 2 1/ v 2 = (9. 81 N/kg)(. 042 m) 2 v = 0. 9078 m/s Work backwards using COM: (. 0065 kg)v = (3. 215 +. 0065 kg)(0. 9078 m/s) v = 449. 9 m/s ≈ 450 m/s

How high above its original position will the block of wood fly? 1. 150 kg . 0062 kg v = 673 m/s . 66 m

So the bullet will impart its momentum to the block: (0. 0062 kg)(673 m/s) = (0. 0062 kg + 1. 150 kg)(v) v = 3. 609 m/s Now energy will be conserved as it flies up in the air: 1/ mv 2 = mgh 2 1/ v 2 = gh 2 1/ (3. 609 m/s)2 = (9. 81 N/kg)h 2 h = 0. 66 m

Whiteboards

A 210. gram air track glider going 0. 190 m/s collides head on with a 420. gram glider going the other way at 0. 750 m/s. The gliders then stick together. What is their post collision speed? How much kinetic energy is lost in the collision? 0. 437 m/s (in the direction of the 420 BTW), 0. 0619 J lost

A 360. gram air track glider going 0. 950 m/s collides with a 410. gram glider going the same way at 0. 150 m/s. The gliders then stick together. What is their post collision speed? How much kinetic energy is lost in the collision? 0. 524 m/s, 0. 0613 J lost

A 3. 20 g bullet going horizontally strikes a 530. g ballistic pendulum at rest and sticks in it, making it swing up to a height of 83. 0 cm. What speed were the block and bullet going just after the collision, and what was the bullet's speed before the collision? 4. 04 m/s, 672 m/s

A 4. 10 g bullet going horizontally at 780. m/s strikes a 870. g ballistic pendulum at rest, and sticks in it making it swing up to some height before going back down. What was the velocity of the bullet and block just after the collision? To what height did the bullet and block combo swing? 3. 66 m/s, 68. 2 cm

A 2. 70 g bullet going 560. m/s vertically upward strikes the bottom of a 480. g block of wood at rest and sticks in the block without emerging. What is the velocity of the bullet and block combo right after the collision? To what height above its original position does the block rise after the collision? 3. 13 m/s, 0. 500 m

A 4. 10 g bullet going straight up at some speed strikes the bottom of a 210. g block of wood at rest, and sticks in it without going through. The bullet and block combo fly 13. 0 m up into the air. What was the post collision speed of the combo, and what was the bullet's original speed? 16. 0 m/s, 834 m/s