A cannon on a railroad car is facing in a direction parallel to the tracks (the

Question

A cannon on a railroad car is facing in a direction parallel to the tracks (the figure below). It fires a 97-kg shell at a speed of 105 m/s (relative to the ground) at an angle of 60.0° above the horizontal. If the cannon plus car have a mass of 5.2 104 kg, what is the recoil speed of the car if it was at rest before the cannon was fired? [Hint: A component of a system's momentum along an axis is conserved if the net external force acting on the system has no component along that axis.]
A.)________ m/s

Explanation / Answer

Sorry I cant read the mass of the car plus the cannon.

so, you have to plug those number in you calculator to find out the final answer.

Now i see lol it 5.2*104 kg


Here:

momentum is P = mv (m stand for mass, v stand for velocity)

where momentum is conserved => P1 = P2

in this situation, Py (momentum in y direction) is not going to affect the final velocity of the car, so it can be ignored.

Pi = Pf

Pxi = mshell*Vxi

=> Pxi = 97 * 105 cos(60o)

Pxf = mcan+cannon*Vxf


=> Pxi = 97 * 105 cos(60o) = Pxf = mcan+cannon*Vxf

=> Vxf = 97 (kg) * 105 (m/s) * cos(60o) / 52000 (kg)

=> Vxf = 0.0979 (m/s)

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