A cannon on a railroad car is facing in a direction parallel to the tracks. It fires a \(98-\mathrm{kg}\) shell at a speed of \(105 \mathrm{m} / \mathrm{s}\) (relative to the ground) at an angle of \(60.0^{\circ}\) above the horizontal. If the cannon plus car have a mass of \(5.0 \times 10^{4} \mathrm{kg},\) 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.]

Initially, before the cannon fires, the total momentum of the system (cannon+car+shell) is 0, as they are at rest.

After the cannon fires, the shell has a momentum, which has a horizontal component. To calculate the horizontal component of the shell's momentum, we will use the formula: momentum_horizontal = mass_shell * velocity_horizontal Where mass_shell is the mass of the shell (98 kg), and velocity_horizontal is the horizontal component of the shell's velocity. To find the horizontal component of the velocity, we use the formula: velocity_horizontal = velocity * cos(angle) Where velocity is the total velocity of the shell (105 m/s), and angle is the angle it makes with the horizontal (60 degrees). Converting the angle to radians: angle_rad = (60 * pi) / 180 Calculating the horizontal velocity: velocity_horizontal = 105 * cos(angle_rad) Substituting the values: momentum_horizontal = 98 * velocity_horizontal

To find the recoil speed of the cannon+car, we will use the conservation of momentum in the horizontal direction. Since the initial momentum was 0, the final momentum of the cannon+car must be equal and opposite to the final momentum of the shell. Using the formula: momentum_cannon+car + momentum_shell = 0 momentum_cannon+car = -momentum_shell We then express the momentum of the cannon+car as the product of the mass and the recoil speed: mass_cannon+car * recoil_speed = -momentum_shell Solving for the recoil speed: recoil_speed = -momentum_shell / mass_cannon+car Substituting the values: recoil_speed = -momentum_horizontal / (5.0 * 10^4)

Now, we can plug in the numerical values to find the recoil speed: angle_rad = (60 * pi) / 180 velocity_horizontal = 105 * cos(angle_rad) momentum_horizontal = 98 * velocity_horizontal recoil_speed = -momentum_horizontal / (5.0 * 10^4) Finally, calculating the recoil speed: recoil_speed ≈ -0.21 m/s The negative sign indicates that the recoil speed of the cannon+car is in the opposite direction of the horizontal component of the shell's velocity.