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kolbaska11 [484]

3 years ago

7

An Arrow (0.5 kg) travels with velocity 20 m/s to the right when it pierces an apple (2 kg) which is initially at rest. After th

e collision, the arrow and the apple are stuck together. Assume that no external forces are present and therefore the momentum for the system is conserved. What is the final velocity (in m/s) of apple and arrow after the collision

1 answer:

sattari [20]3 years ago

6 0

Answer:

4 m/s

Explanation:

From the law of conservation of momentum,

Total momentum before collision = Total momentum after collision

mu+m'u' = V(m+m')...................... Equation 1

Where m = mass of the arrow, u = initial velocity of the arrow, m' = mass of the apple, u' = initial velocity of the apple, V = Final velocity of the apple and the arrow after collision.

make V the subject of the equation

V = (mu+m'u')/(m+m').................... Equation 2

Given: m = 0.5 kg, m' = 2 kg, u = 20 m/s, u' = 0 m/s(initially at rest)

Substitute into equation 2

V = (0.5×20+2×0)/(2+0.5)

V = 10/2.5

V = 4 m/s.

Hence the final velocity of the apple and the arrow after the collision = 4 m/s

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Answer:

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Unknown:

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$F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0$

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b)

$F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3$

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Substituting t as found in eq.(1), we get the equation

$y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}$

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Anonymous

Sep 20, 2010

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d=(1/2)(a)(t^2)+(vi)t

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t=.45 sec

Then you find the horizontal distance traveled by using

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Then you need to find the time of player B by using

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Finally to find player Bs initial horizontal velocity you use the horizontal equation

v=d/t

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