Using the law of conservation of momentum which states that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision.

Let P1A and P1B be the initial momentum of the bodies A and B respectively

Let P2A and P2B be the final momentum of the bodies A and B respectively after collision.

Based on the law:

P1A+P2A = P1B + P2B

Given P1A = 5kgm/s

P2A = 0kgm/s(ball B at rest before collision)

P2A = -2.0kgm/s (negative because it moves in the negative x direction)

P2B = ?

Substituting the values in the equation gives;

5+0 = -2+P2B

5+2 = P2B

P2B = 7kgm/s

3 0

2 years ago

(Mass vs. Weight) HELP PLZ!!

Lina20 [59]

$\huge \boxed { \sf{Answers}}$

c. The weight of an object on the moon will be the same as its weight on Earth. It is false because the weight of an on the moon will be 1/6 th times its weight on Earth.
d. The weight of an object is its mass multiplied by the force of gravity. The statement is false because the formula of weight is mass × acceleration due to gravity, not force of gravity.
e. The mass and weight of an object are the same thing. The statement is false because mass means a body of matter. While weight of an object is its mass multiplied by the force of gravity.
f. The mass of an object is the force of gravity acting upon an object. It is false because it will be the weight of the object not mass.
So, the answers are c, d, e and f.

Hope you could understand.

If you have any query, feel free to ask.

7 0

2 years ago

a projectile is fired in such away that its horizontal range is equal to three times its naximum height.what is the angle of pro

finlep [7]

Answer:

$\theta=53.13^o$

Explanation:

<u>2-D Projectile Motion</u>

In 2-D motion, there are two separate components of the acceleration, velocity and displacement. The horizontal component has zero acceleration, while the acceleration in the vertical direction is always the acceleration due to gravity. The basic formulas for this type of movement are

$V_x=V_{ox}=V_ocos\theta$

$V_y=V_{oy}-gt=V_osin\theta-gt$

$x=V_{ox}t$

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle x_{max}=\frac{2V_{ox}V_{oy}}{g}$

$\displaystyle y_{max}=\frac{V_{oy}^2}{2g}$

The projectile is fired in such a way that its horizontal range is equal to three times its maximum height. We need to find the angle \theta at which the object should be launched. The range is the maximum horizontal distance reached by the projectile, so we establish the base condition:

$x_{max}=3y_{max}$