August 1, 2008
**PHY2053 Discussion**
**Quiz 9 (Chapter 13.1-13.4)**
**Name: UFID:**
*1. (5pts) A 2.00-kg block is attached to a horizontal spring with a spring constant of 600.N/m. The spring is stretched a distance of 0.200 m and the block is released from rest. What is the speed of the block when it passes through the equilibrium position if the coefficient of kinetic friction between the block and the floor is 0.250?
*Applying the work-energy theorem to the initial and final state of the system, we have*
*-μmgA = (1/2)mv²-(1/2)kA² *
⇒ *v = √(kA²/m-2μgA) = √(600×0.2²/2-2×0.25×9.8×0.2) = 3.32 m/s*
**2. (5pts) A 6.00-kg block is attached to a horizontal spring with a force constant of 200 N/m. Another block with a mass of 4.00-kg is pushed against the 6-kg block, compressing the spring by the amount of 0.200 m. The system is then released from rest. The two blocks move back to the equilibrium point together, and then the 4-kg block leaves the 6-kg block. Find the distance between the two blocks when the spring is fully stretched for the first time.
*The velocity of the two blocks at the equilibrium position is given by*
*(1/2)(m*_{1}+m_{2})v² = (1/2)kA² ⇒ v = √{k/(m_{1}+m_{2})}A = √(200/10)×0.2 = 0.894 m/s
*After they pass the equilibrium position, only the 6-kg block moves with simple harmonic motion. The displacement at the fully stretched position and the time interval for the block to reach this position are*
*(1/2)kA’² = (1/2)m*_{1}v² ⇒ A’ = √(m_{1}/k)v = √(6/200)×0.894 = 0.155 m
*t = (1/4)T = (1/4)2π√(m*_{1}/k) = 0.272 s,
*Since the 4-kg block moves at a constant speed during this time interval, the distance between the two blocks is*
*D = x*_{2}-x_{1} = vt-A’ = 0.894×0.272-0.155 = 0.0881 m = 8.81 cm
*3. (5pts) A 15.0-g bullet is fired into and embeds itself in a 600-g block attached to a spring constant of 80.0 N/m and whose mass is negligible. How far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block, and the block slides on a frictionless surface?
*During the collision, the total momentum of the system is conserved. Therefore,*
*mv*_{i} = (m+M)v_{f } ⇒ v_{f} = mv_{i}/(m+M) = 0.015×300/(0.015+0.6) = 7.32 m/s
*After the collision, the mechanical energy of the system is conserved. Thus we have*
*(1/2)(m+M)v*_{f}² = (1/2)kA² ⇒ A = √((m+M)/k)v_{f} = √(0.615/80)×7.32 = 0.642 m
***4. (5pts) A 1.00-kg block is attached to a horizontal spring with a force constant of 40.0 N/m on the frictionless floor of a railroad car. The other end of the spring is attached to the rear wall of the car. Initially the car and the spring-block system are both at rest. Then the car accelerates at 2.00 m/s² until it reaches its final speed of 20.0 m/s. What is the final amplitude of the block? (Hint: The mechanical energy is conserved after the train stops accelerating.)
*When the train is accelerating, the center of the oscillation is the position where the block can be stationary relative to the train. We have*
*ma = -kL*_{0} ⇒ L_{0} = -ma/k
*Thus the amplitude of the oscillation A = 0-L*_{0} = ma/k. The position (measured from unstretched position) and velocity as functions of time are given by
*x+L*_{0} = Acos(ωt) = (ma/k)cos√{(k/m)t}, v = -Aωsin(ωt) = -(ma/k)√(k/m)sin(√(k/m)t)
*It takes 10s (t = v/a = 20/2) the train to reach the final velocity, and after reaching the final velocity the mechanical energy of the block-spring system is conserved. Thus*
*(1/2)kA² = (1/2)kx²+(1/2)mv² (t = 10s) *
⇒ *A² = [ma/k+(ma/k)cos√{(k/m)t}]²+(m/k)[-(ma/k)√(k/m)sin(√(k/m)t)]² *
*= (ma/k)² (1+2 cos√{(k/m)t}+ [cos√{(k/m)t}]²)+(m/k)(ma/k)²(k/m)[ sin(√(k/m)t]²*
*= 2(ma/k)²(1- cos√{(k/m)t}) = 2×(1×2/40)²×(1-cos(√(40/1)×10)) = 4.218×10*^{-4}
⇒ *A = √[2(ma/k)²(1- cos√{(k/m)t})] = √[2×(1×2/40)²×(1-cos(√(40/1)×10))] = 0.0205 m = 2.05 cm* |