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Conservation of energy and momentum

  • Thread starter Lord Dark
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1. Homework Statement

An open container of mass 50.0 kg is rolling to the left at speed of 5.0 m/s on a
frictionless surface. A 15.0 kg box slides down a 37° inclined plane and leaves
the end of the plane with speed of 3.0 m/s. The end of the plane is a vertical
distance of 4.0 m above the bottom of the container (see figure). The box lands
in the container and they roll off together.
a- What are the velocity components of the box just before it lands in the container?
b- What is the magnitude and direction of the final velocity of the container?

2. Homework Equations
U1+K1=U2+K2
Pi=Pf


3. The Attempt at a Solution
a)U1+K1=U2+K2 (U2=0)
mgh=.5mVo^2=.5mv^2 then i get V=9.35 m/s

b)Vx=(15*7.47)-50*5)/65=-2.12 m/s
Vy=(15*-5.627)/65 = -1.3 m/s
so in (b) i get V=2.12 m/s to the left (removed the Vy because the floor is horizontal)
and in (a) is it right to have U2=0 ?? because the question wasn't specific if the container in the ground or not ,, anyway the picture is in the attachment,,are my answers right ?
 

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Doc Al

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a)U1+K1=U2+K2 (U2=0)
mgh=.5mVo^2=.5mv^2 then i get V=9.35 m/s
That's correct for the magnitude, but you need the components.

b)Vx=(15*7.47)-50*5)/65=-2.12 m/s
Vy=(15*-5.627)/65 = -1.3 m/s
so in (b) i get V=2.12 m/s to the left (removed the Vy because the floor is horizontal)
and in (a) is it right to have U2=0 ?? because the question wasn't specific if the container in the ground or not ,, anyway the picture is in the attachment,,are my answers right ?
I don't understand what you did here. Hint: In what direction is momentum conserved during the collision of box and container?
 
121
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That's correct for the magnitude, but you need the components.


I don't understand what you did here. Hint: In what direction is momentum conserved during the collision of box and container?
i got the components and used them in question (b) where Vx=Vcos37 and Vy=Vsin-37 (V=9.35),, the momentum is conserved in the X-axis right ?? because i see that Y-axis just has the block components and there is a floor so the vertical velocity wont count.. is that right ??
 

Doc Al

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i got the components and used them in question (b) where Vx=Vcos37 and Vy=Vsin-37 (V=9.35),,
This is incorrect. The angle 37° applies only to the initial velocity. By the time the block hits the container the angle will be different.
the momentum is conserved in the X-axis right ?? because i see that Y-axis just has the block components and there is a floor so the vertical velocity wont count.. is that right ??
Yes, momentum is only conserved in the x-direction. (There are external forces--from the floor--acting in the y-direction.)
 
121
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This is incorrect. The angle 37° applies only to the initial velocity. By the time the block hits the container the angle will be different.

Yes, momentum is only conserved in the x-direction. (There are external forces--from the floor--acting in the y-direction.)
let see ,, you mean i can distribute Vf into 2 components right .. so the should i devide Vi from the first because Vix=Vfx right ?? so just apply (mgh+.5*m*Viy^2=.5*m*Vfy^2) is this right ??
if it is right I'll get Vix=Vfx=2.4 m/s but Vfy=9 m/s right ?

and from momentum ((2.4*15)-(5*50)=65Vf)) therefore Vf =-3.3 m/s (Vf=3.3 m/s in the container direction) am i right now ? :)
 

Doc Al

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Looks good to me. :approve:
 
121
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cool ,, thanks very much again :)
 

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