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General Physics – ph 201 Name:____________
Final Exam (Ch 1
– 10) Type A
Part I – True or False (5 points each): For questions 1 – 11, state whether each statement is true or false.
Part II – Multiple Choice (5 points each): Choose the one correct answer for each of the following questions that best answers or completes the question.
Net force will be zero, and object will maintain the speed it had just before the counteracting force was applied.
Since f_{s}^{max} = _{s}N = 1.0 . 10N = 10N and F_{T}cos30 = 17.3N, then
∑F_{x} = ma_{x } F_{T}cos30 – f_{s} = ma_{x} since F_{T}cos30 = 17.3N > f_{s}^{max} = 10N, the object will move and accelerate to the right.
According to the free diagrams drawn for each case (above) :
Case1 : N – Mg – F sin = 0 N = Mg + F sin
Case 2: N + Mg sin Mg = 0 N = F sinMg
Case 3: N – Mg cos 0 N = Mg cos
W_{net} = KE = � mv^{2} – � mv_{i}^{2} = � mv^{2}
Since gravity is the only force acting on both ball, and it is a conservative force (its work will not depend on the path taken and only would depend on end points), therefore net work is the same for both ball.
Part III – Problems: Please Show your work clearly and completely for each of the following problems. Partial credit is only given to a work that is shown clearly.
From time interval A to B.
Now from D to E
Now fro C to D
y_{o} = 0 See the diagram for quantities involved
y_{1} = ? Take to roof as origin with downward being positive.
y_{2} = ? To find y_{1}
y = 1.6 v_{1}^{2} = v_{o}^{2} + 2g(y_{1} – y_{o})
v_{o} = 0
v_{1} = ? To find v_{1}, use information of window
v_{2} = ? (we don’t need) y_{2} = y_{1} + v_{1}t + � gt^{2}
t = 0.20 sec
Note: Minimum speed for the block to remain on track at the top is when block is barely in contact with track (i.e. normal force = 0)
a) To find speed at point C:
–mg – N = – mv_{c}^{2}/r (up is +)
N = 0 for minimum speed at A
mg + 0 = mv_{c}^{2}/r gr = v_{c}^{2}
To find minimum h, use conservation of energy
E_{A} = E_{C}
KE_{A} + PE_{A} = KE_{C} + PE_{C}
0 + mgh = � mv_{c}^{2} + mg(2r)
gh = � gr + 2gr
b) To find normal force at B:
N – mg = mv_{B}^{2}/r N = mg + mv_{B}^{2}/r
Use conservation of energy between points A and B to find v_{B}
KE_{A}+ PE_{A} = KE_{B}+ PE_{B} 0 + mg(2h) = � mv_{B}^{2} + 0 v_{B}^{2 } = 4gh
where h = 2.5r v_{B}^{2 } = 4g(2.5r) = 10gr
N = mg + m(10gr)/r
c) –N – mg = – mv_{C}^{2}/r N = mv_{C}^{2}/r – mg
Use conservation of energy between points A and B to find v_{B}
KE_{A}+ PE_{A} = KE_{C}+ PE_{C} 0 + mg(2h) = � mv_{C}^{2} + m(2r)
v_{C}^{2 } = 4gh – 4r v_{C}^{2 } = 4g(2.5r) – 4r v_{C}^{2 } = 6gr
N = m(6gr)/r – mg
d) To find normal force at point D
N – mg = 0
m = 0.0080 kg Isolating the
v_{i} = ? bullet-block
M = 0.25 kg system just
V_{i} = 0 before and just after collision, there will be no external forc
V_{f} = ? acting on the system, hence momentum of the system is
= 0 (horizontal take off) conserved
y_{o} = 1.0 m p_{i} = p_{f}
y = 0 mv_{i} + MV_{i} = mV_{f} + MV_{f} mv_{i} + 0 = (m + M)V_{f}
x_{o} = 0 To find V_{f} which is the same as the horizontal take off
x = 2.0m speed of projectile, we can use the trajectory equation:
g = 9.8 m/s^{2}
V_{f} = 4.4 m/s
Inserting in momentum equation:
0.0080kg v_{i} = (0.0080kg + 0.25kg)(4.4 m/s)
a. Draw a diagram showing the velocity vectors for m_{1} = 3.0 kg (before explosion), m_{2} = 1.2 kg, and m_{3} = 1.8 kg (after explosion).
Note: At top of projectile (before explosion), velocity is horizontal; after collision the velocity of fragments are with respect to the v_{1}. Since the 1.8 kg piece falls vertically then v_{3x} = –v_{1x}, and the 1.2 kg piece must fly horizontally to conserve momentum. Although we have isolated the system, force of gravity is acting on the system, but since the time of explosion is very small, gravity does not act on the system appreciably during the explosion and momentum is conserved.
b. Find the magnitude of the velocity of the 1.2 kg fragment immediately after the explosion. (Answer: 320 m/s)
p_{ix} = p_{f x} m_{1}v_{1x}_{ } = m_{2}v_{2x} + m_{3}v_{3x} (all velocities after collision are with respect to m_{1} before collision).
where v_{1}_{x} = v_{ox} = v_{o}sin 30 and v_{3x} = –v_{1}_{x} = –v_{o}sin 30
3.0 kg (120 m/s sin 30) = –1.8 kg (120 m/s sin 30) + 1.2 kg v_{2x}
v_{2x} = 240 m/s
However, this speed is relative to original piece, so
(v_{2x} )_{ground } = v_{2x} + v_{1x} = 240 m/s + (120 m/s sin 30) v_{2x} = 300 m/s
(v_{3x} )_{ground} = v_{3x} + v_{1x} = – 120 m/s sin 30 – 120 sin 30 (v_{3x} )_{ground} = 0
c. Find the distance between the point of firing and the point at which the 1.2 kg fragment strikes the ground.
Total distance m_{2} has from origin (launching position) is:
x = x_{1} + x_{2} . The time fragments are in the air is 3.2 sec.
x = v_{1x} t + v_{2x} t = 120 m/s sin 30 (3.2 sec) + 300 m/s (3.2 sec)
x = 1152 m
d. Calculate the kinetic energy of the projectile before and after the explosion.
The kinetic energies are to be compared for all velocities with respect to ground:
KE_{i} = � m_{1} v^{2}_{1x} = � (3.0 kg) (120 m/s sin 30)^{2} KE_{i} = 5400 J
KE_{f} = � m_{2}v^{2}_{2x} + = � m_{3}v^{2}_{3x} = � (1.2 kg) (300 m/s)^{2} + 0 KE_{f} = 54000J
In the explosion kinetic energy of the system is not conserved.
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