practice midterm 1 answers

8/6/2019 Practice Midterm 1 Answers

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Solutions to Practice Midterm I (version B)

1) The vectors are,

B 5.6cos33Oi 5.6sin33

Oj

C 4.8sin22Oi 4.8cos22Oj

The vector difference is,

D B C

5.6cos33 4.8sin22Oi 5.6sin33 4.8cos22

Oj

The magnitude D is,

5.6cos33 4.8sin222

5.6sin33 4.8cos222 6.644

ANS.B

2) The vector sum is,

A B C

1 3 1Oi 4 1 1

Oj 1 4 0

Ok

The magnitude is,

1 3 12 4 1 12 1 4 02 7.07

ANS D

3) The scalar product is,

A C 8 6 cos70 40 16.4

ANS. A

4) If the train travels 3.3km and reaches a velocity of 48m/s, we can use the formula to findacceleration,

v12 v0

2 2ax1 x0

482 0 2a3300

a 482

23300.349

ANS C

5) The average acceleration is,

v2 v1t2 t1

30 26

20 103 2800m/s2

ANS.E

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6) ANS A, This is the impossible choice. If an object has constant velocity (that is both thedirection and magnitude are constant) the acceleration must zero and constant.

7) The x-ave. velocity is,

x5. 5 x05. 5 0

7 5. 52 4 5. 5 6 6

5. 5 34.5

The y-ave. velocity is,

y5. 5 y05. 5 0

3 5. 53 3 5. 52 12 5. 5 5 5

5. 5 62.25

ANS B

8) The vertical velocity is given by,

v12 v0

2 2gy1 y0

0 962 2gy1 0

where at the maximum height the velocity v12 is zero. Solving for y1,

y1 962

29. 8 470

ANS E

9) Both bullets feel the same component of gravitation acceleration, g, perpendicular to theincline plane. So the force in perpendicular direction will cause the bullets to veer towardsthe plane with the eqn,

S 12

gt2

where S is the perpendicular distance to the plane. Since S is the same for both bullets, thetime for each bullet to touch the plane must be the same.

ANS E

10) use the dot product relations,

B C cos 6. 19. 3 5. 84. 6

cos 6. 19. 3 5. 84. 6

B C

cos 6. 19. 3 5. 84. 66. 12 9. 32 9. 32 4. 62

arccos6. 19. 3 5. 84. 6

6. 12 5. 82 9. 82 4. 62 1. 2347 radians

1. 2347 180 70. 743deg

ANS C

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11) the acceleration is the same for the ball everywhere and is ay g 9. 8m/s2

ANS A

12) The rocket rose 65m and obtained a positive velocity of 50m/s from ya to yb where we

can assume ya

0 and vy,a

0. Using the relation, vy,b2

vy,a2

2ayyb

ya, we can solve foray

vy,b2

2yb

502

265. Then we use that vy t vy,a ayt and solving for t,

t vy,bay

50502

265

2. 6s

Alternatively, we can more directly solve the problem using, yb ya vbva

2t or 65 50

2t.

So t 6550/2

2. 6s.

ANS A

13) The horizontal velocity is constant and equals 80m/s. The vertical velocity isvy v0y ayt where the y acceleration is due to gravity and equals 9. 8. So at 15s, we havethe velocity magnitude is,

802 600 9. 8 152 460.

ANS E

14) suppose the projectiles are shot with an angle with respect to the horizontal. Wehave,

v0,x v0 cos

v0,y v0 sin

the y position of the projectile is then given by,

y v0,yt12

9. 8t2

This equals zero at y 0 v0,yt12

9. 8t2 or v0,y v0 sin 12

9. 8t. So t is proportional to

sin and so the bigger sin is, the bigger t is. So the maximum time the projectile is in theair when is in closest to 90 deg.

ANS B)

15) Given horizontal speed, v0,x 15.4cos

18030 13. 337 and we can get the initial

vertical speed 15.4sin 180

30 7. 7. So the time t when the arrow reaches zero height is,

yt

0

19

v0 sin30t

1

2 9. 8t

2

or 0

19

7. 7t

4. 9t

2

. We can solve for the time in thisquadratic eqn. Using at2 bx c 0, we have,

t b b2 4ac

2a

7. 7 7. 72 4 4. 9 19

2 4. 9 2. 9058s

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Note the choice 7.7 7.7244.919

24.9would give a negative time which is not chosen.

We can check that, 19 7. 7t 4. 9t2 19 7. 72. 9058 4. 92. 90582 0.

Using this time for the x-position, 13. 337 2. 9058 38. 75m. ANS DRemark; in the Feb 2 tutorial, Prof. Toki, selected the wrong time and incorrectly said ANSB was the answer.

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practice midterm 1 answers

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