A person stands in a stationary canoe and throws a 5.16 kg stone with a velocity of 8.05 m/s at an angle of 31.0° above thehorizontal. The person and canoe have a combined mass of 105 kg. Ignoring air resistance and effects of the water, find thehorizontal recoil velocity (magnitude and direction) of the canoe.

magnitude m/s

direction - --Select-- - opposite the horizontal component of the velocityof the stoneat right angles to the horizontal component ofthe velocity of the stonealong the horizontal component of the velocity ofthe stone

Question

Kayley Mahoney

Physics

19 April, 10:32

A person stands in a stationary canoe and throws a 5.16 kg stone with a velocity of 8.05 m/s at an angle of 31.0° above thehorizontal. The person and canoe have a combined mass of 105 kg. Ignoring air resistance and effects of the water, find thehorizontal recoil velocity (magnitude and direction) of the canoe.

magnitude m/s

direction - --Select-- - opposite the horizontal component of the velocityof the stoneat right angles to the horizontal component ofthe velocity of the stonealong the horizontal component of the velocity ofthe stone

+5

Answers (1)

Know the Answer?

Huynh · Answer 1

Answer: - 0.33909m/s

Explanation:

Mass of stone (Ms) = 5.16kg

Velocity of the stone (Vfs) = 8.05m/s

Angle=31.0 degrees

The person and the canoe (Mc) = 105kg

The conversation of momentum law applied in the horizontal direction gives:

McVfc+MsVfscos (31.0) = 0

Making Vfc subject of the formula:

Vfc=-MsVfscos (31.0) / Mc

= - (5.16kg) (8.00m/s) cos (31.0) / 105kg

= - 0.33909m/s

The magnitude of the velocity is 0.33909m/s. The minus sign indicates the direction opposite the horizontal component of the velocity of the Stone.