Physics Ch 4&amp;5

# Physics Ch 4&amp;5 - Chapter 4 solved problems Physics...

This preview shows pages 1–3. Sign up to view the full content.

Chapter 4 solved problems Physics 1301 Fall 2010, Professor Carlos R. Ordonez 6. Picture the Problem : The motion of the electron is depicted at right. Strategy: Use the given information to independently write the equations of motion in the x and y directions. There will be a pair of equations for the position of the particle (like equation 4-6), except the acceleration will not be the same as g . Use the equations of motion to find the requested time and position information. Let the x direction correspond to horizontal, and the y direction to vertical. Solution: 1. (a) The horizontal motion is character- ized by constant velocity. Apply equation 4-6: ? 0 9 6.20 0 cm 2.95 10 s 2.10 10 cm/s x xx t v 2. (b) Use the time to find the vertical deflection , again using equation 4-6 except substituting for . ag   2 1 00 2 17 2 9 2 1 2 0 cm 0 cm 5.30 10 cm/s (2.95 10 s) 2.31 cm yy y y v t a t Insight: This problem is very much akin to projectile motion, with uniform acceleration in one direction and constant velocity the perpendicular direction. The only difference is the acceleration is 5.30×10 17 cm/s 2 upward instead of 9.81 m/s 2 downward. 17. Picture the Problem : The trajectory of the climber is indicated in the figure at right. Strategy: The 45  direction of motion indicates that, just prior to landing, the climber is falling with a speed equal to his horizontal speed. Use this fact together with equation 4-7 (because the initial velocity is horizontal) find the height difference of the crevasse and the landing point of the climber. Solution: 1. (a) Use the fact that 0 y vv to find the time of flight: 2 7.8 m/s 9.81 m/s 0.80 s y v t g     2. Find the vertical drop during the flight, which is also the height difference between the two sides of    2 22 11 9.81 m/s 0.80 s 3.1 m h y gt 0 v a x y

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
the crevasse: 3. (b) Find the horizontal distance traveled:    7.8 m/s 0.80 s 6.2 m x x v t 4. The climber lands 6.2 – 2.8 m = 3.4 m beyond the far edge of the w = 2.8 m wide crevasse. Insight: The climber impacts the other side of the crevasse at about 25 mi/h (verify this for yourself!). It would be much safer to cross the crevasse with a ladder! 20. Picture the Problem : The pumpkin’s trajectory is depicted in the figure at right. Strategy: The horizontal velocity remains 3.3 m/s throughout the flight of the pumpkin, but the vertical velocity continually increases in the downward direction due to gravity. Use the given time information to find the vertical component of the velocity, and use it together with the horizontal component to find the direction and magnitude of the velocity vector. Equation 4-7 apply because the initial velocity of the pumpkin is horizontal. Solution: 1. (a) Determine the vertical component of the velocity:    2 9.81 m/s 0.75 s 7.4 m/s y y v gt v  2. Find the direction of the velocity vector: 1 7.4 m/s tan 66 or 66 below horizontal 3.3 m/s      3. Find the magnitude of the velocity vector:
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/02/2011 for the course PHYS 1301 taught by Professor Ordonez during the Spring '09 term at University of Houston.

### Page1 / 10

Physics Ch 4&amp;5 - Chapter 4 solved problems Physics...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online