Final practice exam
- An athlete executing a long jump leaves the ground at
a 35 degree angle and travels 7.80m. a) What is the
take off speed? b) If the speed is increased by 5%
how much longer would the jump be?
- An object slides down a 28 degree incline. It is observed
that at the bottom of the ramp the speed is precisely half
of what it would have been without friction. Determine the
coefficient of kinetic friction.
- A 60 kg bungee jumper jumps from a bridge. She is tied
to a bungee cord that is 12m long when unstretched, and
falls a total of 31m. a) Calculate the spring constant
of the bungee cord. Assume that Hooke's law applies.
b) Calculate the maximu acceleration experienced by the
- An 18g rifle bullet traveling 180m/sec burries itself
in a 3.6 kg pendulum hanging on a 2.8m long string. The
mass of the string can be ignored. Determine the
horizontal component of the pedulum's displacement.
- A rod of mass m=1kg and length l=2m is pivoted on a
frictionless hinge at one end. The rod is held at rest
horizontally and the released. Determine the speed
of the rod's tip at the moment when it reaches the
vertical position. (Note: The moment of inertia of the
rod relative to an axis through the tip is I=1/3*M*l^2.)
- v_y0=v_0*sin(35). v_x0=v_0*cos(35). T=2v_y0/g.
Then v_0=sqrt(g*x/(2*sin(35)*cos(35))=9 m/sec.
x proportional to v^2. x_new = 7.8m*1.05^2 = 8.6m.
- F_T=sin(28)*m*g, F_N=cos(28)*m*g. a=(F_T-mu*F_N)/m=
(sin(28)-mu*cos(28))*g. Use v=sqrt(2*a*x).
Then v(mu)=1/2*v(mu=0) implies that sin(28)-mu*cos(28)
=1/4*sin(28) or mu=3*tan(28)/4=0.4.
- energy conservation m*g*h=1/2*k*x^2=1/2*k*(h-l)^2.
Then k=2*m*g*h/(h-l)^2=101 N/m.
Acceleration a=F/m-g=k*(h-l)/m-g=22.2 m/sec^2.
- Momentum conservation p_i=m_b*v_b=p_f=(m_b+m_p)*v_p.
Then v_p=m_b/(m_b+m_p)*v_b=0.9 m/sec.
Energy conservation 1/2*(m_b+m_p)*v_p^2=(m_b+m_p)
*g*l*(1-cos(alpha)). Then cos(alpha)=1-v_p^2/(2*g*l)
=0.985 and x=l*sin(a)=0.48 m.
- Energy conservation: mg*(l/2)=1/2*I*w^2=1/2*(1/3*m*l^2)*w^2.
Then w=sqrt(3*g/l) and v=l*w=sqrt(3*g*l)=7.7 m/sec.
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