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Tutorial 6 Solutions For Exercises: Solution

This document contains solutions to physics exercises involving calculations of momentum and velocity. Exercise 4 involves calculating the momentum of a garbage truck and finding the speed needed for a trash can to have the same momentum. Exercise 5 calculates the average force applied to a bullet accelerated to 600 m/s in 2 ms. Exercise 6 finds the final velocity when two moving train cars collide. Exercise 7 considers an elastic collision between two approaching satellites and finds their final relative velocity.

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131 views2 pages

Tutorial 6 Solutions For Exercises: Solution

This document contains solutions to physics exercises involving calculations of momentum and velocity. Exercise 4 involves calculating the momentum of a garbage truck and finding the speed needed for a trash can to have the same momentum. Exercise 5 calculates the average force applied to a bullet accelerated to 600 m/s in 2 ms. Exercise 6 finds the final velocity when two moving train cars collide. Exercise 7 considers an elastic collision between two approaching satellites and finds their final relative velocity.

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Tenielle
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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1.

10902-Physics 1A
Physics Division

Tutorial 6 Solutions for Exercises

4. (a) What is the momentum of a garbage truck that is 1.20 104 kg and is moving at
10.0 m/s ? (b) At what speed would an 8.00-kg trash can have the same momentum as
the truck?

Solution:

(a) p  mv  1.20  104 kg 30.0 m/s  3.60  105 kg  m/s


p 3.60  105 kg  m/s
(b) p  mv  v    4.50  104 m/s
m 8.00 kg

5. A bullet is accelerated down the barrel of a gun by hot gases produced in the combustion
of gun powder. What is the average force exerted on a 0.0300-kg bullet to accelerate it
to a speed of 600 m/s in a time of 2.00 ms (milliseconds)?

Solution:
p mv 0.0300 kg  600 m/s
net F     9.00  103 N
t t 2.00  103 s

6. Train cars are coupled together by being bumped into one another. Suppose two loaded
train cars are moving toward one another, the first having a mass of 150,000 kg and a
velocity of 0.300 m/s, and the second having a mass of 110,000 kg and a velocity of
 0.120 m/s . (The minus indicates direction of motion.) What is their final velocity?
Solution:

m1v1  m2 v2  (m1  m2 )v' ,


m v  m2 v2 150,000kg 0.300 m/s  110,000kg  0.120 m/s
v'  1 1   0.122 m/s
m1  m2 150,000kg  110,000kg

7. Two manned satellites approach one another at a relative speed of 0.250 m/s, intending
to dock. The first has a mass of 4.00  103 kg , and the second a mass of 7.50 103 kg . If
the two satellites collide elastically rather than dock, what is their final relative velocity?
Solution:
In a one-dimensional elastic collision of two objects, the momentum and kinetic
energies before and after the collision are conserved:
2 2 2
m1 (v1  v1' )  m2 (v2'  v2 ) and m1 (v1  v1 )  m2 ( v 2  v1 )
2 ' '
Solving the two equations gives the result: v1  v2  (v1'  v'2 )
So, the relative velocity of approach equals the relative velocity of recession = 0.25 m/s

Answers Only to Chapter 8 [Linear momentum]-Highlighted in Course Outline

2. (a) 1.20 x 108 kg, (b) 1.21 x 103

4. Exercise Question #4

6. 1.78 x 1029 kg·m/s

7. Exercise Question #5

9. (a) 2.40 x 103 N, (b) force same but in opposite direction.

12. F = 6.67 x 106 N

17. 2.10 x 103 N

23. Exercise Question #6

24. v = 0.272 m/s

27. v = 22.4 m/s

29. Exercise Question #7

30. Vgoalie = 0.150 m/s, Vpuck = -34.9 m/s

32. (a) V = 1.78 /s (b) -267 J

36. (a) V = 0.182 m/s, (b) 8.52 x 103 J

42. V = 24.8 m/s

43. V= 1.07 m/s

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