A cannon ball of mass mB = 36kg is fired from a cannon of mass mC 162kg which is

Question

A cannon ball of mass mB = 36kg is fired from a cannon of mass mC 162kg which is on a flat horizontal surface. The cannon is free to recoil horizontally without rolling friction, and stays in contact with the surface during the firing The cannon's barrel is orientated at an angle of 20°above the horizontal. The cannonball exits the barrel with a speed of v109m/s WITH RESPECT TO the barrel. B.C Cannon-ball (B) Cannon (C) (a) What is the firing speed of the cannonball with repsect to a stationary observer on the flat surface? | (b)What is the launch angle of the cannonball with respect to a stationary observer on the flat surface (in degrees)?

Explanation / Answer

(a) Horizontal component of the velocity of the cannon ball, vx1 = 109*cos20 = 102.4 m/s

Vertical component of the velocity of the cannon ball, vy1 = 109*sin20 = 37.3 m/s

Now apply conservation of momentum to know the recoil velocity of the cannon -

0 = 36*vx1 + 162*v2

=> v2 = -36*102.4/162 = - 22.7 m/s

Therefore, x-component of the firing speed of the cannon ball with respect to a stationary observer

= 102.4 - 22.7 = 79.7 m/s

y-component of the firing speed of the cannon ball with respect to a stationary observer = 37.3 m/s

So, the total speed of the cannon ball with respect to an stationary observer = sqrt(79.7^2 + 37.3^2)

= 88 m/s

(b) Launch angle with respect to an stationary observer = tan^-1 (37.3/79.7) = 25.1 deg.

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