Ten quantities that are pure numbers

Let's examine over the next few questions an NFL kick as described in the 3rd Law video. The announcer claimed that the ball is

hichkok12 [17]

Answer:

8451.62109367671 Newtons

Explanation:

$1\ kg=1\times 9.8066500286389=9.8066500286389\ N$

$1 \lb=2.2046226218488\ kg$

It is known that

$1\ lbs=\dfrac{9.8066500286389}{2.2046226218488}=4.4482216282508\ N$

So,

$1900\ lbs=1900\times 4.4482216282508\\\Rightarrow 1900\ lbs=8451.6210936765\ N$

The force in Newtons is 8451.6210936765 Newtons

4 0

4 years ago

A body A of mass 1.5kg, travelling along the positive x-axis with speed 4.5m/s, collides with

Lena [83]

REFER TO THE IMAGES for the SOLUTIONS TO YOUR PROBLEM. Each step will be explained here.

When you solve for velocities before or after collision, you need to remember the law of conservation of moment which can be expressed through this formula:

BEFORE AFTER
m1v1+m2v2 = m1v1 + m2v2

This basically means, the sum of momentum of 2 objects BEFORE collision is equal to the same 2 objects AFTER collision.

The type of collision we have in your case is a 2D collision, where there is a gliding collision or they touch at an angle. So when you solve these type of problems, you have to consider the x and y components of motion. It makes things easier if you make a table like in the image to sort out your components.

STEP 1: COMPUTE FOR MOMENTUM BEFORE COLLISION for each OBJECT involved.
To solve for momentum, the formula is mass x velocity or mv:

STEP 1a: Body A: The problem states that before collision Body A is moving along the positive X-axis so the velocity will be +4.5 m/s. Notice that the velocity of the y component is 0 m/s. This is because BODY A is moving along the x-axis, with no mention that it deviated from it.

STEP 1b: Body B: Body B is at rest before collision, that is why it is not moving at all, which means both x and y components are equal to 0.

STEP 1c: Get the sum of all X components and the Sum of Y components.

STEP 2: COMPUTE FOR MOMENTUM AFTER COLLISION for each OBJECT involved.

Step 2a: BODY A: Notice that we now have an angle. hence the cos and sin. We do this because we are breaking or decomposing the diagonal velocity into its x and y component. To get the x-component you get the cos of the angle and multiply it to the momentum of the diagonal or overall velocity. For y-component, instead of cos, you get the sin.

Step 2b: BODY B: Here we have unknowns, which we will derive later on. In this step, just plug in what you know into the formula.

Step 2c: We already know the x and y momentum of the objects BEFORE collision and the law of conversation of momentum states that the momentum AFTER is the same. With this total we can move onto the next step.

STEP 3: Solving for the X and y component of the velocity of BODY B AFTER collision.

Step 3a: Using the formula given in the image, we plug in what we know first. We know the momentum of the BODY A already, so we can put it into the equation. We also know the sum of both momenta and we put that into the equation too. Now all we do is derive the formula. DO NOT FORGET THAT WE ARE TO USE ONLY X COMPONENTS.

Step 3b: is the same as the previous step but instead, we use Y COMPONENTS only.

STEP 4: Combining X and Y components to get the resultant velocity:
For this step you need to remember the Pythagorean theorem. This is applied here because when you draw a free body diagram of the velocities, it creates a right triangle where :
the hypotenuse represents the final velocity
the opposite side represents the y-component and;
the adjacent represents the x-component.

Refer to the image for the solution.

STEP 5: Solving for the angle at which BODY B is moving:
For this step you need to remember SOH CAH TOA to find the angle at which BODY B is moving. You already have all the components you need, including the hypotheses. You can use any of the functions, and they should come up with the same approximation.

FINAL ANSWER: BODY B was moving at 1.35 m/s, 21 degrees above the x-axis.