Thanks for the help guys... another.?

Question

a 3.0kg mass sliding on a frictionless surface into a 1.0 *** and a 2.0 mass. after the explosion the velocity of the 1kg mass is 5.0m/s due north, and that of the 2kg mass is 7.0m/s, 45 degrees south of east. what was the original speed and direction of the 3kg mass?

Answer 1

I'll measure all angles counterclockwise from east.

Conservation of momentum

E direction
(3kg)(initial speed) cos theta = 2kg(7m/s)cos(-45)

N direction
(3kg)(initial speed) sin theta = 1kg (5m/s)

Two equations; two unknowns--initial speed and direction. Solve for them.

Divide one equation by another to cancel speed.

tan theta = 1kg (5m/s) / (2kg(7m/s)cos(-45))

Plug and chug

then plug theta back in to get the speed.

Answer 2

The statement didn't come through. I will assume the 3 kg mass explodes into two masses, with directions and velocities as described.

Velocity is a vector quantity. I will resolve the East/West and Nort South components first

The 1 kg has only Notherly direction and the 2 kg has sin(45)*7 southerly

The center of mass is moving
1*5-2*sin(45)*7
=4.9 kg*m/s South

The Easterly component is
2*cos(45)*7
=9.9 kg*m/s East

to compute the original velocity, divide each quantity by 3kg
and find the magnitude
=SQRT((4.9/3)^2+(9.9/3)^2)
magnitude is 3.7 m/s

The direction is
Atan(4.9/9.9) South of East
or 26 degrees South of East

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