How Does Jumping on and off a Sled Affect Kinetic Energy and Speed?

[logglypop]

A 30kg child moving at 4m/s jump onto a 50kg sled that is initially at rest on a long frictionless,horizontal.
a) find the speed after the child jumps onto the sled
b)find Kinetic Energy of the child after the child jumps onto the sled

After coasting at constant speed for a short time, the child jumps off the sled in such away that she is at rest with respect to the ice
c) find the speed of the sled after the child jumps off
d) find the kinetic energy of the child-sled system when the child is at rest on the ice
e)Compare the Kinetic energy that were determined in part b and d . If the energy greater in d that b, where did the increase come from? . If the energy is less in d than b , where did the energy go.


a)30*4+50*0=v(30+50)
v=1.5

b)1/2(30+50) *1.5^2
KE=90

c)1.5(80)=1.5(30)+50v
v=1.5

d)0

e) d have less energy than b, the energy transform to heat

 

[Astronuc]

c) find the speed of the sled after the child jumps off
d) find the kinetic energy of the child-sled system when the child is at rest on the ice

c)1.5(80)=1.5(30)+50v
v=1.5

d)0

I don't believe these are correct. The child jumps of so that his/her velocity is zero with respect to the ground, and that is only possible if the child transfers momentum to the sled, which means the sled must increase in velocity.

At rest the child's KE is zero, certainly, but what about the sleds KE. The sled has some velocity.

 

[logglypop]

I don't believe these are correct. The child jumps of so that his/her velocity is zero with respect to the ground, and that is only possible if the child transfers momentum to the sled, which means the sled must increase in velocity.

At rest the child's KE is zero, certainly, but what about the sleds KE. The sled has some velocity.

c) 1.5(80)=1.5(0)+50v
v=2.4m/s

Im not really understand wat u talking about
I think you mean this
d)0+1/2*50*2.4^2
KE=144N ??
Correct me if i am wrong

 

[PhanthomJay]

A 30kg child moving at 4m/s jump onto a 50kg sled that is initially at rest on a long frictionless,horizontal.
a) find the speed after the child jumps onto the sled
b)find Kinetic Energy of the child after the child jumps onto the sled

After coasting at constant speed for a short time, the child jumps off the sled in such away that she is at rest with respect to the ice
c) find the speed of the sled after the child jumps off
d) find the kinetic energy of the child-sled system when the child is at rest on the ice
e)Compare the Kinetic energy that were determined in part b and d . If the energy greater in d that b, where did the increase come from? . If the energy is less in d than b , where did the energy go.


a)30*4+50*0=v(30+50)
v=1.5

looks good

b)1/2(30+50) *1.5^2
KE=90

this is the new KE of the child-sled system which i think is what you meant to say in part b statement?

c)1.5(80)=1.5(30)+50v
v=1.5

all speeds should be referenced to the ice, this is incorrect

d)0

this asks about the child/sled system. The sled is still moving, isn't it, so it can't be zero, can it?

e) d have less energy than b, the energy transform to heat

redo part d and reanswer this part.

 

[Astronuc]

The kinetic energy in the system is due to the sled after the child is at rest.

The KE is in units of Joules (J), if m is in kg and v in m/s.

The answers appear to be correct.


Here is a reference on inelastic collisions - http://hyperphysics.phy-astr.gsu.edu/hbase/inecol.html

 

[logglypop]

here i redo c d again
c) 1/2(30+50)1.5^2= 0+ 1/2(50)v^2
v=1.9m/s

d) the child at rest so his KE=0
0+ 1/2*50*1.9^2 = 90J

 

[PhanthomJay]

here i redo c d again
c) 1/2(30+50)1.5^2= 0+ 1/2(50)v^2
v=1.9m/s

d) the child at rest so his KE=0
0+ 1/2*50*1.9^2 = 90J

whoa, wait, your last set of answers were correct...as Astronuc noted...I posted in between responses and was referring to your original answers. Your revised ones were correct as noted. Just answer part e.

 

[logglypop]

e) d has more answer than b, and i don't really know where the increase speed come from