# Impulsive Motion

1. Apr 13, 2005

### VSCCEGR

Problem:

The initial velocity of block A is 9m/s. Knowing that uk=.3 Determine the time for the block to reach zero velocity. When (theta)=20 deg.

Here is my work so far.

mv_1+Imp_1,2=mv_2

Imp_1,2=Ff(t)+sin(theta)W(t)
Ff=Cos(theta)uk(mg)
W=mg
Imp=9.81cos(20).3m+9.81sin(20)m
Imp=6.12mt

9m+6.12mt=0
t=1.47s

This answer is Wrong It Should Be .963s

What Have I Missed?

2. Apr 13, 2005

### marlon

Strange, i get the exact same answer 1.47

marlon

3. Apr 13, 2005

### marlon

$$v_x = v_i -g(0.3cos(20)+sin(20))t$$
$$v_y = 0$$

X is along the incline and y perpendicular to it. I get the same answer using a different approach. Are you sure it's not 1.47 ?

marlon

4. Apr 13, 2005

### VSCCEGR

Yes, I did place the positive x-axis in the direction of movement, but it really should not matter because I am using the Principle of Impulse and Momentum.
(a) part was finding the time taken if (theta)=0deg. (3.06s I got this one right)
So I know my method works. There is just something I am missing.

The problem is from Beer and Johnston;
VECTOR MECHANICS for ENGINEERS:STATICS and DYNAMICS 7th Ed.
problem 13.124
For anyone who knows the book. (They may have made a mistake but i doubt it.)

Last edited: Apr 13, 2005
5. Apr 13, 2005

### Spectre5

The answer for part (a) is .306 or is is 3.06?

6. Apr 13, 2005

### VSCCEGR

7. Apr 13, 2005

### Spectre5

OK, so for 0 degrees I get 3.06 s

and for 20 degrees I get 1.47 s

I think the book is wrong

8. Apr 13, 2005

### marlon

If you take the angle to be zero, my method also gives 3.06, just plug in that value into the equation for v_x

marlon

ps : i think there must be a mistake here. Perhaps someone else can double check this problem ?

Dexter, Doc Al, what do you get ??

9. Apr 14, 2005

### Staff: Mentor

1.47s is correct. Using "impulse/momentum" is a perfectly OK way to solve this problem; it is equivalent to finding the acceleration and then using kinematics.

Last edited: Apr 14, 2005