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Finding Recoil Velocity 
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#1
Feb607, 11:56 AM

P: 16

1. The problem statement, all variables and given/known data
Oliver has a mass of 59.1 kg and can throw a 558.0 g rock with a speed of 19.8 m/s. What would Oliver's recoil speed be if he were on an icy surface? 2. Relevant equations Momentum is conserved in an isolated system (when Fnet=0 or no external forces are acting on the system) p=mv mass(object 1)*initial velocity (object 1) + mass(object2)*initial velocity (object 2)= Mass (object 1)*final velocity (object 1) + mass (object 2)*final velocity (object 2) 3. The attempt at a solution m (oliver)= 59.1 kg, m(rock)= 0.588 kg, initial velocity = 19.8 m/s p(oliver+ rock)= mv p= (59.1+0.588)(19.8) =1181.8 kg m/s (seems quite too large> unreasonable?) on ice (frictionless i assume) no net external forces so Fnet=0 i don't understand what recoil velocity means? Is it the rebound velocity? would be equal and opposite? ***Any help would be appreciated **** 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution 


#2
Feb607, 12:10 PM

Mentor
P: 41,311

Realize that Oliver and his rock are originally at rest on the ice. So what is the original momentum of the sytem?
Also, "recoil velocity" is the speed that Oliver will move back with (recoil) after he throws that rock. (If he throws the rock East, he will recoil to the West.) 


#3
Feb607, 12:13 PM

P: 2,047

Recoil velocity refers to velocity with which Oliver would be moving AFTER he throws the ball. Take a look at the equation you posted and just substitute the values. 


#4
Feb607, 01:02 PM

P: 16

Finding Recoil Velocity
Would the orginal momentum be zero
so would i use 0 as the momentum and find velocity from there? i am so confused also it says that if oliver throws the rock he would throw it at 19.8 m/s so is that the velocity at the instant he throws it? i seem to be more confused..? 


#5
Feb607, 01:20 PM

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P: 41,311




#6
Feb607, 01:22 PM

P: 2,047




#7
Feb607, 01:23 PM

P: 2,047

Okay, I'm late by only 2 minutes this time. :D



#8
Feb607, 04:45 PM

P: 16

Its seems that since that 19.8 m/s is the velocity oliver will continue with, hence that would be the recoil velocity?
But i tried that its not the answer. Ughhh... I am gonna think about this question later since i should really study for my bio midterm. 


#9
Feb607, 06:17 PM

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#10
Feb607, 07:18 PM

P: 16

(mass of oliver)* (oliver's velocity)= (mass of rock) (velocity of rock) (59.1)(v)= (0.588) (19.8) v= 0.588*19.8/59.1 v=0.196? 


#11
Feb707, 02:10 AM

P: 2,047

From the origianl equation: 0 = m_{s}v_{sf} + m_{o}v_{of} (s refers to stone and o to Oliver, f  the final velocity) Therfore, v_{of} = (m_{s}v_{sf})/m_{o} The minus sign is important because you are solving for velocity, and not just speed. The sign says that Oliver would move in the opposite direction with a speed of 0.197m/s. 


#12
Jun1510, 12:06 AM

P: 2

Can someone please help me solve this problem?
A 60kg teenager on inline skates intially at rest holds a 0.5kg ball. The teenager throws the ball horizontally at a speed of 12m/s relative to the ground and recoils backwards. Ignoring friction, what is the recoil speed of the skater????? 


#13
Jun1510, 07:05 AM

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P: 41,311




#14
Jun1510, 11:09 PM

P: 2

Can someone please help me solve this problem?
A 60kg teenager on inline skates intially at rest holds a 0.5kg ball. The teenager throws the ball horizontally at a speed of 12m/s relative to the ground and recoils backwards. Ignoring friction, what is the recoil speed of the skater????? Can you please show me in steps???? and what formula do I use????? 


#15
Jun1610, 05:14 AM

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P: 41,311

As far as which formula to use: See my hint! And read the thread that you chose to hijack! (It's practically the same problem.) If you have no idea at all about conservation of momentum, read up in your text. 


#16
Jun312, 02:11 PM

P: 1

Use m1*initial velocity1 + m2*initial velocity2= m1*final velocity1+ m2*final velocity2
m1= Oliver m2= The Rock Oliver has no initial velocity so initial velocity1 is gone. The rock's finals velocity will be zero, so final velocity2 is gone. You're left with: m1+m2*initial velocity2=m1*final velocity1+m2 You should get his recoil velocity from that. If you don't then I'm wrong..and it wouldn't be the first time lol. (59.1)+(.558)(19.8)=(59.1)*final velocity2+.558 


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