Trying to find HOW to get the velocity.

  • #1
Homework Statement:
Suppose a 15.0 g dart is fired into a wooden block that is mounted on wheels with a total
mass of 10.0 kg. The time required for the block to travel a distance of 45.0 cm is
measured after the dart lodges into the block. This can easily be accomplished with a
pair of photocells and an electric clock. If the coefficient of friction between the
wheels and floor is 0.1435 and the measured time to come to a stop is .80 seconds, what
is the muzzle velocity of the dart? (assume no drag force on the dart before it hits the
block).
Relevant Equations:
x= vt + 1/2 at^2
m1v1=m2v2
This is a practice problem so I know that the answer is 750 m/.s. Not totally sure what to do with the friction or if any of my listed equations are relevant here but here goes what I've tried:

.45 kg = v(0.8 sec) + 1/2(-9.8 *0.1435) * (0.8)^2

.45 kg = v(0.8) + 1/2(-1.4063) * 0.64
.45=v(0.8) - 0.450015
0.900016= V(0.8)
V does not equal 1.12 m/s
 

Answers and Replies

  • #2
caz
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You only calculated the initial velocity of the block with the dart in it. You still need to calculate initial velocity of dart. You have also written some units incorrectly, but it does not effect the math. The force of friction is -.1435mg so the acceleration is -.1435g.
 
  • #3
haruspex
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There are some oddities with the question.
We are given the "coefficient of friction". Assuming the wheels are free to rotate, the friction will be static, and its coefficient irrelevant. So presumably they mean the rolling resistance coefficient.
It says the time to travel a stated distance is measured, but at that point in the text does not specify that time.
Later, it gives the time to come to a stop. Are we supposed to assume these refer to the same event, i.e. it comes to a stop at the given distance?
It seems so, since on that basis we can calculate the rolling resistance, and it is indeed the value given. So whoever set the question must have been aware it was redundant information. Unusual.

.45 kg = v(0.8 sec) + 1/2(-9.8 *0.1435) * (0.8)^2
.45cm.

While I typed all that, @caz beat me to pointing out your error.
 
Last edited:
  • #4
You only calculated the initial velocity of the block with the dart in it. You still need to calculate initial velocity of dart. You have also written some units incorrectly, but it does not effect the math. The force of friction is -.1435mg so the acceleration is -.1435g.
I see so then I would just apply the equation: m1v1=m2v2 and solve for V1?
 
  • #5
caz
Gold Member
498
470
Yes.
 

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