August 29, 2008
PHY2054 Discussion-Fall ‘08
Warm Up Quiz (Physics 1 Review)
Name:
UFID:
**1. (0pts) A 1.80-m-tall basketball player stands on the floor 5.00 m from the basket. If he/she
shoots the ball at an angle of 45.0˚ to the horizontal, at what initial speed must he/she throw the
basketball so that it goes through the hoop without striking the backboard? The height of the basket
is 3.05 m.
The time interval it takes the ball to reach the basket is expressed as
Δx = v0cosθt ⇒ t = Δx/(v0cosθ)
When the ball reaches the basket, the vertical displacement should be Δy = 3.05-1.8 = 1.25 m.
Plugging the expression for t into the kinematics equation for Δy, we solve the equation for v0:
Δy = v0sinθt-(1/2)gt² = v0sinθ[Δx/(v0cosθ)]-(1/2)g[Δx/(v0cosθ)]² = Δxtanθ-(1/2)g[Δx/(v0cosθ)]2
⇒ v02 = gΔx²/[2cos2θ(Δxtanθ-Δy)]
v0 = √[g/{2(Δxtanθ-Δy)}]×Δx/cosθ = √[9.8/{2(5tan45˚-1.25)}]×5/cos45˚ = 8.08 m/s
***2. (0pts) A 6.00-kg copper block on a 60.0˚ incline is attached to a light string which passes over
a pulley at the top of the incline and then connected to a 2.00-kg aluminum block on a horizontal
surface. The pulley has a mass of 2.00 kg and it can be treated as a solid cylinder (I = (1/2)MR2)
with a radius of 0.200 m, and it rotates without slipping. The coefficient of kinetic friction between
the copper block and the incline is 0.300, and that between the aluminum block and the horizontal
surface is 0.400. Find the acceleration of the two blocks.
If we take the positive directions in the directions of motions/rotations, the relation between the
accelerations of two blocks and the angular acceleration of the pulley is given by a1 = a2 = αR (=
a). Applying Newton’s 2nd law to the copper block, the aluminum block and the pulley, we have
m1a = m1gsinθ-μ1m1gcosθ-T1
m2a = T2-μ2m2g
Iα = (T1-T2)R ⇒ (1/2)MR²(a/R) = (T1-T2)R ⇒ (1/2)Ma = T1-T2
We add these three equations together and solve it for a:
(m1+m2+(1/2)M)a = m1gsinθ-μ1m1gcosθ-μ2m2g
⇒ a = [(m1sinθ-μ1m1cosθ-μ2m2)g]/( m1+m2+(1/2)M)
= [(6sin60˚-0.3×6cos60˚-0.4×2)9.8]/(6+2+0.5×2) = 3.81 m/s²
**3. (0pts) A 2.00-kg pendulum bob is suspended by a light string with a length of 0.500 m. The
pendulum is released from an initial angle of 45.0˚ to the vertical. At the bottom of the swing, the
pendulum bob makes an elastic collision with a 1.00-kg block resting on a horizontal surface. How
far does the block slide on the surface if the coefficient of kinetic friction between the block and
surface is 0.250?
Before the pendulum hits the block, the mechanical energy of the pendulum is conserved. Thus the
speed of the pendulum when it hits the block is
m1gh = (1/2)m1v12 ⇒ v1 = √(2gh) = √(2gL(1-cosθ)) = √[2×9.8×0.5(1-cos45˚)] = 1.69 m/s
Since the collision is elastic, the relative speed and the total momentum are both conserved. The
velocity of the block just after the collision is given by
v1i = v2f-v1f ⇒ v1f = v2f-v1i
m1v1i = m1v1f+m2v2f ⇒ m1v1i = m1(v2f-v1i)+m2v2f
⇒ v2f = 2 m1v1i /(m1+m2) = 2×2×1.69/(2+1) = 2.25 m/s
After the collision, the kinetic energy is of the block is lost due to the friction. Thus we have
-μm2gΔx = -(1/2)m2v2f² ⇒ Δx = v2f²/(2μg) = 2.25²/(2×0.25×9.8) = 1.03 m
*4. (0pts) The distance between two successive crests of a transverse wave is 0.800 m. Six crests
pass a given point along the direction of travel every 15.0 s. Calculate the wave speed.
The number of waves that passes a given point in one second is
f = 6/15 = 0.4 (Hz)
Each wave has a wavelength of 0.8 m, thus the speed (the distance the wave travels in one second) is
v = fλ = 0.4×0.8 = 0.32 m/s