Collision of cballs of clay help

[oldunion]

All of these problems are the last of a series of ten, which i cannot solve. I've given extensive efforts but am now nearly out of time (12pm est). if anone could be of help please feel free.


A 20 g ball of clay traveling east at 2m/s collides with a 30 g ball of clay traveling 30 degrees south of west at 1m/s. what is the speed of the resulting blob, m/s?

Dan is gliding on his skateboard at 2.00m/s. He suddenly jumps backward off the skateboard, kicking the skateboard forward at 7.00m/s(as measured by an observer on the ground). Dan's mass is 60.0 kg and the skateboard's mass is 6.00kg. how fast is dan going as his feet hit the ground

A 10.0g ball of clay traveling east at 7.00m/s collides and sticks together with a 10.0g ball of clay traveling north at 5.50m/s. what is the speed of the resulting clay

A spaceship of mass 2.3×10^6kg is cruising at a speed of 6×10^6m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 4.8×10^5kg, is blown straight backward with a speed of 2.3×10^6m/s. A second piece, with mass 7.6×10^5kg, continues forward at 1.3×10^6m/s.

 

[arildno]

Do you agree that 1&3 is basically the same problem?
What do you know of inelastic collisions?

 

[oldunion]

i just solved the spaceship problem.

they are all momentum problems, and i feel as though i understand them all but something isn't quite right. and i know of ineleastic collisions.

for the skateboard question, i did 60kg(Vf)=6kg*2m/s+60kg*2m/s

for blob of clay question (without degrees) i did (.01kg+.03kg)Vf=.02kg*3m/s+.03kg*2m/s

 

[arildno]

The "blob of clay" questions need you to regard velocities as VECTORS, not scalars as you have done

 

[oldunion]

could you be more specific, do i find x and y components and then add? I am almost positive I am doing it correctly. for the first clay question... (.01kg +.01kg)Vf=.07kgm/s+.055kgm/s final answer of 6.25 is incorrect

 

[arildno]

Right, in so far as you mean to say you add componentwise.
For example, the first one:
Let the unit vector "east" be \vec{i} and north \vec{j}
Then, the unit vector which makes 30 degrees angle to the unit "west" vector in the southern direction is: \vec{t}=-\cos(30)\vec{i}-\sin(30)\vec{j}
Hence, the final velocity, \vec{v}_{f} fulfills:
(m_{1}+m_{2})\vec{v}_{f}=m_{1}v_{1}\vec{i}+m_{2}v_{2}(-\cos(30)\vec{i}-\sin(30)\vec{j})
m_{1}=20g, v_{1}=2\frac{m}{s}, m_{2}=30g, v_{2}=1\frac{m}{s}

 

[oldunion]

thank you so much, but what about the skateboarder, i can't figure it out.

edit: that equation does not work. but I am out of time. thanks for the help

 

[arildno]

Do you know what speed is?
It is not the same as velocity, I thought you knew the difference.

 

[oldunion]

their desired answer is in the form of m/s, the correct answer is .410. how did they come to this?

 

[arildno]

Well, you must calculate the speed from the velocity and plug in the numbers.
To continue with the one I started with, the final speed, ||\vec{v}_{f}||
satisfies:
||\vec{v}_{f}||=\frac{1}{m_{1}+m_{2}}\sqrt{(m_{1}v_{1}-m_{2}v_{2}\cos(30))^{2}+m_{2}^{2}v_{2}^{2}\sin^{2}(30)}=\frac{1}{20+30}\sqrt{(20*2-30*1*\frac{\sqrt{3}}{2})^{2}+30^{2}*1^{2}*\frac{1}{4}}