Calculating Speed in an Inelastic Collision

[jessguy]

Hi everyone!

I've been on the question for a while and I'm not quite sure where I'm going wrong. If someone could help me out, I'd really appreciate it!

Problem
Two ice-fishermen are driving trucks across a frozen and frictionless lake. Truck 1 (m1 = 1190kg) is traveling with a speed of 13.2m/s in a direction of 43.0degrees SOUTH of EAST. Truck 2 (m2 = 1000kg) is traveling due NORTH with a speed of 20.1m/s. The trucks collide and lock into a single unit. What is the speed of the joined trucks immediately after the collision?

Homework Equations

equation for inelastic collision...

Vf = (m1V1i + m2V2i) / (m1 + m2)

The Attempt at a Solution

I have tried it a couple of ways so far, and neither have worked.


First, I tried just plugging in the velocities without using the angle.

Vf = (1190kg*13.2m/s + 1000kg*20.1m/s) / (1190kg + 1000kg)
= 16.4 m/s (wrong answer!)


Then I tried incorporating the y-coordinate angle however I'm not sure if I did this properly.

Vf = (1190kg*13.2m/s(sin43) + 1000kg*20.1m/s) / (1190kg + 1000kg)
= 14.1 m/s (wrong again!)




If you could offer me any help, it would be greatly appreciated!
:)

 

[LowlyPion]

Welcome to PF.

Keep in mind that momentum is a vector. When they join won't the result be a vector addition?

The equation you are using would be appropriate if the motion was all in 1 direction.

 

[tiny-tim]

Welcome to PF!

Two ice-fishermen are driving trucks across a frozen and frictionless lake. Truck 1 (m1 = 1190kg) is traveling with a speed of 13.2m/s in a direction of 43.0degrees SOUTH of EAST. Truck 2 (m2 = 1000kg) is traveling due NORTH with a speed of 20.1m/s. The trucks collide and lock into a single unit. What is the speed of the joined trucks immediately after the collision?
…
Then I tried incorporating the y-coordinate angle however I'm not sure if I did this properly.

Vf = (1190kg*13.2m/s(sin43) + 1000kg*20.1m/s) / (1190kg + 1000kg)
= 14.1 m/s (wrong again!)

Hi jessguy! Welcome to PF!

Momentum is a vector, so you must use vector addition.

Add the x-coordinates to get V_x, then the y-coordinates to get V_y (that's easy in this case!), and use Pythagoras' theorem to find the total speed, V.

 

[jessguy]

hm.. ok. so is the mass not needed for this part of the question? (there is a second part to this question, you see)

This is what I've tried after receiving your feedback..

Vx = 13.2m/s*cos43 = 9.65 m/s

Vy = 20.1m/s - 13.2m/s*sin43 = 29.1 m/s

Vf = sqrt((9.65m/s)^2 + (29.1m/s)^2)
= 14.7 m/s

Not the correct answer, once again.

I understand that I need to use the x and y components of velocities. Am I suppose to use the masses and if so, how?


thanks again

 

[tiny-tim]

I understand that I need to use the x and y components of velocities. Am I suppose to use the masses and if so, how?

Yes … use your original equation, Vf = (m1V1i + m2V2i) / (m1 + m2).

This is a vector equation, but like all vector equations, it also works for coordinates in one direction.

 

[jessguy]

ooook! perfect! I got the answer afterwards!Thanks so much :)