Dynamics troubles

[F for Freedom]

Hey guys, I'm not in any physics class but I found this Advanced Level Physics textbook (its pretty old, printed in 1987) and found it really interesting. So I'm reading along and doing the problems at the ends of each section. Conceptually I understand (most of) everything but I am having a few troubles with some of the problems.

Thanks in advance for any pointers in the right direction for any of the problems.

Note: I know no Latex. vf = final velocity; vi = initial velocity; % = theta. Feel free to question my syntax so I can clarify any ambiguity.

1. Question:
In a nuclear collision, an alphaparticle A of mass 4 units is incident with a velocity v on a stationary helium nucleus B of 4 mass units. After collision, A moves in the direction BC with a velocity v/2, where BC makes an angle of 60 degrees with the initial direction AB, and the helium nucleus moves along BD [there is a diagram where AB is the line/direction where A hits B; BC is the line/direction where A shoots off to going northeast; and BD is where B shoots off to headed southeast].
Calculate the velocity of rebound of the helium nucleus alogn BD and the angle made with the direction AB.

My Work:
mv + mv = mv + mv

x is the velocity of B; v is the velocity of A.
X-direction:
4v = 4(v/2)cos60 + 4xcos%
4v - 2vcos60 = 4xcos%
3v = 4xcos%

Y-direction:
0 = 4(v/2)sin30 + 4(-x)sin%
2vsin30 = 4xsin%

(2vsin30) / (3v) = (4xsine%) / (4xcos%)
(2/3)sind30 = tan%
% = 18 degrees.

I stopped here because...

Book's Answer: .9v, 30 degrees.


2. Question:
A large cardboard box of mass .75kg is pushed across a horizontal floor by a force of 4.5N. The motion of the box is opposed by (i) a frictional force of 1.5N between the box and the floor, and (ii) an air resistance force kv^2, where k = 6.0 * 10^(-2) kg/m and v is the speed of teh box in m/s.
Calculate maximum values for (a) the acceleration of the box, and (b) its speed.

My Work:
a.) F = ma
4.5 - 1.5 = .75a (why isn't it 4.5 - 1.5 - kv^2 = .75a ?)
3 = .75a
a = 4 m/s^s

b.) F = ma
4.5 - kv^2 = .75a
kv^2 = 1.5

Thats all I have.

Book's Answer: (a) 4 m/s^2 (b)7.1 m/s



I know this is probably really time consuming for anyone, so thanks again for any help.

[Doc Al]

Y-direction:
0 = 4(v/2)sin30 + 4(-x)sin%
2vsin30 = 4xsin%
That should be sin60, not sin30.

[Doc Al]

My Work:
a.) F = ma
4.5 - 1.5 = .75a (why isn't it 4.5 - 1.5 - kv^2 = .75a ?)
3 = .75a
a = 4 m/s^s
Hint: For what value of speed will the net force be maximum?

b.) F = ma
4.5 - kv^2 = .75a
kv^2 = 1.5
Hint: The maximum speed will be achieved when the box stops accelerating.

[Andrew Mason]

1. Question:
In a nuclear collision, an alphaparticle A of mass 4 units is incident with a velocity v on a stationary helium nucleus B of 4 mass units. After collision, A moves in the direction BC with a velocity v/2, where BC makes an angle of 60 degrees with the initial direction AB, and the helium nucleus moves along BD [there is a diagram where AB is the line/direction where A hits B; BC is the line/direction where A shoots off to going northeast; and BD is where B shoots off to headed southeast].
Calculate the velocity of rebound of the helium nucleus alogn BD and the angle made with the direction AB.This is also an elastic collision so energy is conserved. Because energy is conserved, v_{f\alpha}^2 + v_{He}^2 = v_{i\alpha}^2.

This also tells you that the two particles move at 90 degrees to each other!

The vertical components of final momentum must sum to _____?

The horizontal components of final momentum must sum to _____?

AM