# Glancing Collisons

A 50.0 Kg Object is moving east at an unknown velocity when it collides with a 60.0 Kg stationary object. After the collsion, the 50.0 Kg object is travelling at a velocity of 6.0 m/s 50.0° N of E, and the 60.0 Kg object is travelling at a velocity of 6.3 m/s 38° S of E.

What was the velocity of the 50.0 Kg object before the collision?

Can someone please explain to me how this question is solved. Thank-you.

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First of all, you are supposed to post your own solution before you ask for help.

What is your specific problem with this question? Do you know that the change in linear momentum of a system on which no external force acts, is zero? Can you think of applying this principle to this problem?

Sorry i didnt post my solution earlier, but i did not no what to do after this:

a=1st object
b=2nd Object
Ma=50.0 Kg
Va=?
Mb=60.0 Kg
Vb=0 m/s
Va'= 6.0 m/s 50.0° N or E
Vb'=6.3 m/s 38° S of E

Pa + Pb = Pa' + Pb'
Pa= Pa' + Pb'

Now from here what do you do in order to solve for Pb'?

Hi

You have to resolve the velocities in mutually perpendicular directions (say x and y) and write either one vector equation for the conservation of linear momentum or (which is the same thing really) two separate equations correspdonding to momenta along the two chosen axes. That will give you the equations to solve for the velocity components.

Like this

$$P_{A,x} + P_{B,x} = P'_{A,x} + P'_{B,x}$$
$$P_{A,y} + P_{B,y} = P'_{A,y} + P'_{B,y}$$

Cheers
Vivek