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3 years ago

One of the rectangular components of the velocity 40ms^1 is 20 ms^1. find the other components

Physics

1 answer:

Zielflug [23.3K]3 years ago

7 0

Answer:

The other component is 17.32m/s.

For more information 

lemme so you how to solve it!

given velocity v = 20 m/s

let the given rectangular component is

cos somponent, so 10 = v cos x

so, other component will be sin component

i.e., v sin x

or, 10 = 20 cos x

or, cos x = 1/2

and as sin^2 x + cos^2 x = 1

or, 1 - cos^2 x = sin^2 x

or, 1 - (1/2)^2 = sin^2 x

so, sin x = sqrt( 1 - 1/4)

or, sin x = sqrt(3/4)

so, rectangular component is

20 sin x = 20 sqrt(3/4)

= 10 sqrt3

= 17.32 m/s

Explanation:

hope to help you (:

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First let us sort out our given:Mass should be in kg to get the proper answer. Now let's assign m1 as the mass of the dart and m2 as the mass of the block.
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$5.5g$ x $\frac{1kg}{1000g}= 0.0055kg$

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$22.6g$ x $\frac{1kg}{1000g}= 0.0226kg$

So now we settled that we can set our given as:
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Now you can see that we have 2 unknowns: v1i and vf. We need the vf to solve for the initial velocity of the dart or object 1. We have other given to consider, so we can make use of that to get our missing vf.

Now, vf is the horizontal velocity after the collision. We do this by first using the equations for projectiles considering that we have an x and y dimension to consider. We use the y dimension to get the x.

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We can use the UAM equations to solve for the time in the y-dimension (vertical) to get the horizontal velocity.

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$1.5 = (0)t+\frac{1}{2} (9.8)t^{2}$

$1.5 = \frac{1}{2} (9.8)t^{2}$

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$\sqrt{0.3061} = \sqrt{t^{2}$

$0.553s = t$

Now using this, we can get the horizontal (x-dimension) velocity using the formula:
$v_{x} =d_{x}t$ and our given earlier for the horizontal distance is 2.5m and we solved for time 0.553s. Let's put that into our equation:
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Now we finally have our vf or velocity after the collision. Now let's get back to the equation.

$m_{1}v_{1i}+m_{2}v_{2i}=(m_{1}+m_{2})v_{f}$

From this we can derive the equation for v1i by isolating it.

$v_{1i}= \frac{((m_{1}+m_{2})v_{f})-(m_{2}v_{2i})}{m_{1}}$

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$v_{1i}= \frac{((0.0055kg+0.0226kg)4.52m/s)-((0.0226kg)0m/s)}{0.0055kg}$

$v_{1i}= \frac{(0.0281kg)4.52m/s)}{0.0055kg}$

$v_{1i}= \frac{0.127012kg.m/s}{0.0055kg}$

$v_{1i}= 23.09m/s$

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