Question

A quarterback is set up to throw the football to a receiver who is running with a constant velocity ~vr directly away from the quarterback and is now a distance D away from the quarterback. The quarterback estimates that the ball must be thrown at an angle θ to the horizontal and the receiver must catch the ball a time interval tc after it is thrown. Assume the ball is thrown and caught at the same height y = 0 and the horizontal position of the quarterback is x = 0. Use g as the magnitude of the gravitational acceleration.

(a) Find v0y, the vertical component of the velocity of the ball when the quarterback releases it. Express your answer in terms of tc and g. [6]

(b) Find v0x, the horizontal component of the velocity of the ball when the quarterback releases it. Express your answer in terms of D, tc and vr. [6]

(c) Find the speed v0 with which the quarterback must throw the ball. [4]

(d) Find the angle θ above horizontal at which the quarterback must throw the ball. [4]

2. A spring-loaded toy cannon is used to shoot a ball of mass m = 1 kg straight up in the air. The spring has the spring constant k = 334 N/m. If the spring is compressed a distance of 50.0 cm from its equilibrium position y = 0 and then released, the ball reaches a maximum height hmax (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the cannon. Assume that all movement occurs in a straight line up and down along the y axis.

(a) Is mechanical energy conserved? [2]

(b) Use conservation of energy to find vc, the velocity of the ball when it emerges from the cannon (i.e. at the spring’s equilibrium position y = 0.) [10]

(c) Now assume that the ball touches the cannon wall and the frictional force exerting the ball is F = 10N. Use the conservation of energy to find vc. [8]

3. Collision: (a) A ball of mass m moving with velocity ~vi strikes a vertical wall. The angle between the linear velocity vector of the ball and the wall before collision is θi . The duration of the collision is ∆t, and the collision is completely elastic. Neglect friction.

i. Use the laws of conservation of energy and momentum to prove that the angle θf that the velocity vector of the ball makes with the wall after the collision is equal to θi . [10]

ii. Find an expression for the average force F exerted by the wall on the ball during the collision, in terms of the variables given.

4. (a) Two blocks are suspended by a massless cord over a pulley. The mass of block B is twice the block A. The cord does not slip or stretch. There is no friction between the pulley and the axle. Find the net torque in the system. [10]

(b) A simple wheel has the form of a solid cylinder of radius r with a mass m uniformly distributed throughout its volume. The wheel is pivoted on a stationary axle through the axis of the cylinder and rotates about an axle at a constant angular speed. The wheel rotates n full revolution in a time interval t. Find the kinetic energy K of the rotating wheel. Express your answer in terms of m, r, n and, t. Assume that moment of inertia I = 0.5mr2 .

Answer #1

a)

consider the motion along the vertical direction :

v_{oy} = initial velocity along the vertical
direction

a_{y} = acceleration due to gravity = - g

y = vertical displacement = 0

t_{c} = total time of travel

Using the kinematics equation

y = v_{oy} t + (0.5) a_{y} t^{2}

0 = v_{oy} t_{c} + (0.5) (- g)
t_{c}^{2}

v_{oy} = (0.5) (g) t_{c}

b)

Consider the motion along the horizontal direction

v_{ox} = component of velocity along the horizontal
direction

x_{o} = initial position of receiver = D

v_{r} = speed of receiver

x = horizontal displacement of the ball

t_{c} = total time of travel

horizontal displacement of the ball is given as

x = x_{o} + v_{r} t_{c}

Also

x = v_{ox} t_{c}

hence

v_{ox} t_{c} = x_{o} + v_{r}
t_{c}

v_{ox} t_{c} = D + v_{r}
t_{c}

v_{ox} = (D/ t_{c} ) + v_{r}

C)

speed is given using Pythagorean theorem as

v_{o} = sqrt(v_{ox}^{2} +
v_{oy}^{2})

v_{o} = sqrt(((D/ t_{c} ) + v_{r}
)^{2} + ((0.5) (g) t_{c})^{2})

D)

Angle is given as

= tan^{-1}(v_{oy}/v_{ox}) =
tan^{-1}(((0.5) (g) t_{c})/((D/ t_{c} ) +
v_{r} ))

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