Projectiles, Momentum, and work

blader324

Hey, i just wanted to check my answers with somebody on this question. It's an even number so i can't check with the back of the book, and my final is tomorrow. Here's the question:

A soccer ball is kicked 56 degress above the ground at an inital velocity of 12m/s. the mass of the soccer ball is 75 grams. what is the maximum height of the ball? I got 5.05 meters

A car is driving up a hill that's 40 degress above the horizontal. Mass of the car is 2000 kg, initial velocity of the car is 22m/s. The mew between the car and the hill is (.36). At 5.05 meters, the soccer ball collides inelastically with the car. What is the velocity of the car and the soccer ball together? I got 21.2m/s

Lastly. ignoring the soccer ball being attached to the car: the hypotenuse of the hill is 17m. How much work does friction remove from the system? I got -211837 Joules

They're not that hard...but i was wondering if anyone could help me out. THANKS!

learningphysics

I also get 5.05 for that first part. But I get something different for the 2nd and 3rd parts... can you explain how you got the 2nd and 3rd parts?

blader324

okay, so for the second part (momentum) that the initial velocity of the ball is zero because it's at the top of it's motion as it hits the car...so then, all i had was the mass and velocity of the car= (added masses)(added velocities)

blader324

for the third part (the work) i just drew a free body diagram of the car...and found that the normal force is 34614 N. i found the force of friction with that information (friction force=12461.04), and then i used the work equation. and got -211837 Joules

learningphysics

You didn't convert the grams to kg.

Also, remember that the ball has a horizontal velocity... I think the trick for this problem is that momentum is conserved along the slope. So you'd get the component of the momentum of the soccer ball along the slope... then use conservation of momentum along the slope.

The thing is that the mass of the soccer ball is just 0.075 and makes little difference... the velocity I get after the collision is very slightly less than 22m/s. 21.999m/s. It's a strange question...

blader324

okay, so then would i use the 12m/s as the initial velocity for the ball for the momentum equation? because...the ball hits the car at the TOP of its motion, and doesn't that mean that the ball's velocity is zero?

learningphysics

I get 2000*9.8*cos40 = 15014N for the normal force

learningphysics

The vertical velocity is 0. But the horizontal velocity the same as what it was at the beginning which is 12*cos56 = 6.71m/s

blader324

okay...so the velocity doesn't change, and i just stick with the original component? as for the FBD, since the car isn't technically "accelerating" does that mean that i don't need to do a net force in the y direction=mass X acceleration in the y direction?

learningphysics

By y direction do you mean perpendicular to the slope? Isn't that how you calculated the normal force? acceleration would be 0 so net force is 0.

blader324

okay...so since the car isn't speeding up or slowing down, then the net forces in the y direction would be equal to zero?

learningphysics

Yes that's correct.

blader324

sweet...so then for the momentum problem i got 21.99 as the final velocity...and then the work....i got the normal force as 15014.47 and then the work done by friction as -91888.56 Joules...is that right?

learningphysics

That's what I get also.

blader324

THANK YOU SO MUCH! ugh...you saved me learningphysics...MUCHAS GRACIAS!!!

learningphysics

You're welcome.