Question

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

Answer 1

The recoil velocity of the canoe is calculated using conservation of momentum. The horizontal component of the stone's velocity is first determined, and the stone's momentum is then used to find the canoe's recoil. The direction of recoil is opposite to the stone's horizontal velocity component.

Step-by-step explanation:

The recoil velocity of a stationary object after projecting an object can be calculated using the principle of conservation of momentum.

Since the person and the canoe are initially motionless, the total momentum before the stone is thrown must be equal to the total momentum afterwards.

Step-by-Step Explanation

1. Calculate the horizontal component of the stone's velocity using the given angle and the initial velocity (vcos(31.0°)).
2. Find the stone's momentum in the horizontal direction by multiplying its mass by the horizontal component of its velocity.
3. Apply conservation of momentum: The momentum imparted to the stone will be equal in magnitude and opposite in direction to the momentum of the person and canoe.
4. Calculate the recoil velocity of the canoe by dividing the stone's horizontal momentum by the mass of the person and canoe combined.

The direction of the canoe's recoil will be opposite to the stone's horizontal component of velocity since momentum is conserved and no external horizontal forces are acting on the system.

Answer 2

Answer: -0.33909m/s

Step-by-step 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.