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Aleksandr-060686 [28]
3 years ago
15

Question 9

Physics
1 answer:
egoroff_w [7]3 years ago
6 0

Answer:

C

Explanation:

F=ma

given solution

v=12m/s a=v/t

s=6 sec =12m/s÷6sec

=2m/s^2 then we get acceleration now we will find the mass. first derive the the formula of mass by crisis cross then you will get this formula which is m=F/a

=36÷2

= 18

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a 5.5 g dart is fired into a block of wood with a mass of 22.6 g. the wood block is initially at rest on a 1.5 m tall post. afte
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<span>From the problem alone we can say that the dart and the block of wood combined into a single object moving together at the end. With that clue we know that the collision is an inelastic collision. The formula of an inelastic collision is:

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

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So now we settled that we can set our given as:
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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. 
</span>

dy = -1.5 m 

a = 9.8m/s^2

viy = 0 (take note that the initial vertical velocity is 0)

t = ?

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

dy = v_{iy}t +  \frac{1}{2} at^{2}</span>

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

<span>1.5 = \frac{1}{2} (9.8)t^{2}

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

\frac{(3)}{9.8}=t^{2}

\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:
v_{x} =d_{x}t
v_{x} =(2.5m)(0.553s) 
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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}

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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

The initial speed of the dart is 23.09 m/s or 23.10 m/s.</span>
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