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3 years ago

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Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t

hey collide head–on. If the resulting velocity of Car A after the collision is –10 m/s, what is the velocity of Car Z after the collision?

1 answer:

salantis [7]3 years ago

5 0

Explanation:

Mass of bumper cars, $m_1=m_2=110\ kg$

Initial speed of car A, $u_1=8\ m/s$

Initial speed of car Z, $u_2=-10\ m/s$

Final speed of car A after the collision, $v_1=-10\ m/s$

We need to find the velocity of car Z after the collision. Let it is equal to $v_2$ . Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$110\times 8+110\times (-10)=110\times (-10)+110v_2$

$v_2=\dfrac{-1320}{110}\ m/s$

$v_2=-12\ m/s$

So, the velocity of car Z after the collision is (-12 m/s). Hence, this is the required solution.

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When working with these types of equations, you just need to know your kinematics equations. For projectiles launched at an angle, you will need to first break down the initial velocity (Vi) into its horizontal (x) and vertical (y) components.

*<em>Now remember this, if you are solving for something in the horizontal movement always use only x-components. When solving for vertical movements, always use y-components. </em>

Let's move on to breaking down the initial velocity into both y and x components.

Viy = SinΘVi = (Sin30°)(19.5m/s) = <em>9.75 m/s</em>

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(i) Maximum height above the ground

Remember that the object was thrown 1.50m above the ground. So we save that for later. First we need to solve for the maximum height above the horizontal, or the point where it was thrown.

The kinematics equation you will use is:

$Vf^{2} = Vi^{2}+2ad$

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We will derive our displacement from this equation. And you will come up with this:

$d = \dfrac{Vf^{2}-Vi^{2}}{2a}$

Again, remember that we are looking for a vertical component or y-component because we are looking for HEIGHT. So we use this plugging in vertical values only.

Vf at maximum height is always 0m/s because at maximum height, objects stop. Also because gravity is a downwards force you will use -9.8m/s².

Vfy = 0 m/s a = -9.8m/s² Viy = 9.75m/s

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$dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}$

$dy = \dfrac{0^{2}-(9.75m/s)^{2}}{2(-9.8m/s^{2})}$

$dy = \dfrac{-95.0625m^{2}/s^{2}}{-19.6m/s^{2}}$

$dy = 4.85m$

So from the point it was thrown, it reached a height of 4.85m. Now we add that to the height it was thrown to get the MAXIMUM HEIGHT <em>ABOVE THE GROUND.</em>

4.85m + 1.50m = 6.35m

(ii) Speed before it strikes the ground. (Vf=resultant velocity)

Okay, so here we need to consider a couple of things. To get the VF we need to first figure out the final velocities of both the x and y components. We are combining them to get the resultant velocity.

Vfx = horizontal velocity = Initial horizontal velocity (Vix). This is because gravity is not acting upon the horizontal movement so it remains constant.

Vfy = ?

VF =?

We need to solve this, again, using the same formula, but this time, you need to consider we are moving downwards now. So this time, instead of Vfy being 0 m/s, Viy is now 0 m/s. This is because it started moving from rest.

$Vfy^{2} = Viy^{2}+2ad$

$Vfy^{2} = 0m/s^{2}+2(9.8m/s^{2}(6.35m)$

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$Vfy= 11.16m/s$

OKAY! We are at our last step. Now to get the resultant velocity, we apply the Pythagorean theorem.

$Vf^{2} = Vfx^{2} + Vfy^{2}$

$\sqrt{Vf^{2}} = \sqrt{(16.89m/s)^{2}+(11.16m/s)^{2}}$

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