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An interpreted illustration of a physical scenario related to a Newton's laws problem. In the scene, a blue sphere representing a body of mass 5kg is moving due south at a high speed, illustrated by a blur effect and directional arrow. It collides with a smaller, stationary yellow sphere representing a body of mass 3kg. Post-collision, both spheres are shown moving together in the same southward direction, slower than the original blue sphere, as suggested by a lighter blur effect and a smaller arrow. The scene is devoid of textual elements.

A body of mass is 5kg moving with a velocity of 20m due south hits a stationary body of mass is 3kg if they move together after collision with a velocity due south what is d value of l

Given:

M1 = 5kg, V1 = - 20m/s.
M2 = 3kg, V2 = 0.
V3 = Velocity of M1 and M2 after colliding.

Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V3.
5 *(- 20) + 3 * 0 = 5V3 + 3V3,
8V3 = -100,
V3 = -12.5 m/s = 12.5 m/s, South.

Well, I do not know what

"d value of l "
means but
final momentum south = initial momentum south
so
5 * 20 = (5+3) * new v south = momentum before and after
100 = 8 v
v = 100/8 m/s south

Thanks

Well, the value of "l" is missing from your question. However, given that the bodies move together after the collision with a velocity due south, I can suggest a few humorous interpretations:

1. If "l" stands for "laughter," then the value of "l" is priceless because this situation would be quite entertaining to watch!

2. If "l" stands for "late," well, it seems like both bodies are heading south in a hurry, so expect them to be a bit "l"ate to their destination!

3. If "l" stands for "luck," then the value of "l" might indicate that luck is on their side, as the bodies managed to stick together and continue moving south without any problems.

Remember, humor doesn't always have to be serious!

To find the value of "d," we will need to use the law of conservation of linear momentum. According to this law, the total momentum before the collision is equal to the total momentum after the collision.

The momentum of an object is given by the product of its mass and velocity. Therefore, the momentum of the 5kg body before the collision is calculated as:

Momentum1 = Mass1 * Velocity1

Momentum1 = 5kg * 20m/s = 100 kg·m/s due south

Similarly, the momentum of the 3kg body before the collision is zero since it is stationary.

Now, let's consider the momentum after the collision. Since the two bodies move together after the collision, we can add their momenta together:

Momentum2 = (Mass1 + Mass2) * Velocity2

Momentum2 = (5kg + 3kg) * Velocity2

Momentum2 = 8kg * Velocity2

Since we want the velocity after the collision to be due south, we can assume the velocity is negative (-Velocity2).

Now, we can equate the initial momentum (Momentum1) to the final momentum (Momentum2):

100 kg·m/s = -8kg * Velocity2

To find the value of "d," we need to solve for Velocity2. Rearranging the equation, we have:

Velocity2 = -100 kg·m/s / 8kg

Velocity2 = -12.5 m/s

Since we assumed the velocity after the collision to be negative and in the south direction, the value of "d" is 12.5 m/s.