Question

1. When you throw a pebble straight up with initial speed V, it reaches a maximum height H with no air resistance. At what speed should you throw it up vertically so it will go twice as high?

a. Insufficient information

b. 18 V

c. V sqrt(2)

d. 4 V

e. 8 V

2. A 60.0-kg man stands at one end of a 20.0-kg uniform 10.0-m long board. How far from the man is the center of mass (or center of gravity) of the man-board system?

a. 1.25m

b. 5m

c. 9m

d. 2.5m

Answer 1

solution

1)(c)Vsqrt(2)

we know at maximum height vertical speed becomes zero (Vyf=0)

by using the motion equation , final speed

Vyf^2=Voy^2+2(-g)H

where Voy=intial vertical speed =V

0=V^2-2gH

H=V^2/2g

now throw by a speed V' to reach the ball at the height of 2H

Vyf^2=V'^2+2(-g)(2H)

0=V'^2-4gH

V'^2=4gH=4g(V^2/2g)=2V^2

V'=V*sqrt(2)..........................answer

(2)a. 1.25m

to find the location of the center-of-mass point of a system. you first choose a location that will be defined as x=0. I chose the center of the board as x=o, so the person is standing at x=5 (end of the board). the formula for finding the center of mass is, to sum up, all the different masses (or center of mass of each smaller part of the system) times their position, and then divide that sum in the total mass. in our case

xcm = (20*0 + 60*5) / (60 + 20) = 300 / 80 = 3.75

that means the total center of mass is at x=3.75. the distance from that point to where the man is standin is

d = | xman - xcm | = | 5 - 3.75 | = 1.25.

the correct answerr is a.