Question

A 0.17 kg arrow with a velocity of 14 m/s to
the west strikes and pierces the center of a 3.0
kg target.
What is the final velocity of the combined
mass?
Answer in units of m/s

Answer 1

Please show work and don't round.

Answer 2

I got 0.75

and 0.751
both wrong though.

Answer 3

KE of arrow = 0.5mv^2 = 0.5(0.17)(14^2) = 16.66J

When the arrow pierces the target, assuming no loss of energy (as heat, sound, etc) and using the principle of conversion of energy, we have:

KE of combined mass = initial KE of arrow = 16.66J
Using KE = 0.5mv^2, we have:

Velocity of combined mass = \[\sqrt{2(16.66)/(3+0.17)}\] = 3.242 m/s.

Answer 4

When I typed this answer in it said it was not correct.

Answer 5

Did you type 3.242? Or was there an exact number of significant figures required for this question? Or did the question require you to answer in exact form?

Answer 6

exact form

Answer 7

If it's in exact form, you'd have to leave it as the following:

\[\sqrt{33.32/3.17}\]
And please ensure that the units are in m/s.

Answer 8

this is a momentum question right? well, it can be worked out using momentum anyway, so yeah, this is how i would do this.

\[p_{arrow} = mv\]
\[p_{arrow} = .17 \times 14\]
\[p _{arrow} = 2.38 kgm/s\]
\[p _{target} = 0\]
now we add the weights and their momentums

\[p _{total} = 2.38 kgm/s\]
\[m _{total} = 0.17kg+3.0kg\]
\[m _{total} = 3.17kg\]
rearrange the equation to find v
\[v = \frac{ p }{ m }\]
\[v = \frac{ 2.38 }{ 3.17 }\]
\[p = .7507886435 ms ^{-1}\]

Answer 9

@bennyboy489: This is interesting - you use the principle of conservation of linear momentum to solve it while I use the principle of conservation of energy to solve it, and we get totally different solutions (even though both are not entirely correct as there is some loss of energy and momentum in the process). Bobthebad, perhaps you might wish to try inputting bennyboy489's answer (2.38/3.17) into the system (this is the exact, unrounded form) & see if this is the correct answer.

Answer 10

yeah, i know, it does seem strange, but that is the way i would look at this problem, it would be interesting to see, .75ms-1 does seem about right, because the arrow is hitting an object 17 times its size

Answer 11

@bennyboy489: Yes, I think your answer's correct since it seems that there is a greater loss of energy than momentum in the case of an inelastic collision, hence it is more meaningful to use the principle of conservation of linear momentum than the principle of conservation of energy.

Answer 12

by the sounds of it, this question is probably proposed in a frictionless environment, so all energy is conserved in the collision, provided there is no other external force