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Agata [3.3K]
3 years ago
5

Two objects have the same momentum. Do the velocities of these objects necessarily have (a) the same directions and (b) the same

magnitudes? Give your reasoning in each case.
Physics
1 answer:
NeX [460]3 years ago
3 0

Answer:

(a) They must have same direction

(b) It is not necessary for them to have same magnitudes

Explanation:

(a)

Momentum is a vector quantity. It is the product of mass (scalar) and velocity (vector). Thus, if the direction of velocity is changed, then as a result the direction of momentum will also change or its magnitude or component in the same direction will change. Hence, for the two objects to have same momentum, the directions of their velocities must also be the same.

(b)

Since, the momentum is product of velocity and mass. It is possible that two bodies of different masses with different velocities might have same momentum, provided the direction of their velocities is same.

For example, take a body of mass 4 kg moving with speed 5 m/s. It will have a momentum of 20 N.s. Now, consider another body of mass 2 kg, moving with speed 10 m/s. It will also have a momentum of 20 N.s.

Thus, it is not necessary for two objects to have same magnitude of velocity to have same momentum.

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Two massless bags contain identical bricks, each brick having a mass M. Initially, each bag contains four bricks, and the bags m
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Answer: F_{2}=\frac{3}{4}F_{1}

Explanation:

According to Newton's law of universal gravitation:

F=G\frac{m_{1}m_{2}}{r^2}

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G is the universal gravitation constant.

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r is the distance between both bodies

In this case we have two situations:

1) Two bags with masses 4M and 4M mutually exerting a gravitational attraction F_{1} on each other:

F_{1}=G\frac{(4M)(4M)}{r^2}   (1)

F_{1}=G\frac{16M^2}{r^2}   (2)

F_{1}=16\frac{GM^2}{r^2}   (3)

2) Two bags with masses 2M and 6M mutually exerting a gravitational attraction F_{2} on each other (assuming the distance between both bags is the same as situation 1):

F_{2}=G\frac{(2M)(6M)}{r^2}   (4)

F_{2}=G\frac{12M^2}{r^2}   (5)

F_{2}=12\frac{GM^2}{r^2}   (6)

Now, if we isolate \frac{GM^2}{r^2} from (3):

\frac{F_{1}}{16}=\frac{GM^2}{r^2}   (7)

Substituting \frac{GM^2}{r^2}  found in (7) in (6):

F_{2}=12(\frac{F_{1}}{16})   (8)

F_{2}=\frac{12}{16}F_{1}   (9)

Simplifying, we finally get the expression for F_{2}  in terms of F_{1} :

F_{2}=\frac{3}{4}F_{1}  

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