Conservation of momentum and concept of center of mass.

[Satvik Pandey]

Homework Statement

A frog sits on the end of a long board of length L=5m . The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board . What is the minimum take-off speed (in s.i. Units) i.e.,relative to ground 'v' that allows the frog to do the trick ? The board and the frog have equal masses.

Homework Equations

The Attempt at a Solution

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As no external force act in the horizontal direction on the system so horizontal velocity of the frog is equal to the velocity of the plank (their directions are opposite). And also the position of the CoM of the system at final position and initial position should be same.

Now , the CoM of the plank will be displaced by length ##5/2##.

So ##t=\frac{5}{2V_{x}}##.....(1)

In this time period the from will jump and land on the plank.

So ##t=\frac{V_{y}}{5}##.....(2)

Using eq(1) and (2) we get

##V_{y}V_{x}=\frac{25}{2}##...(3)

Let the net velocity of the frog be ##v_{0}##

So ##{ V }_{ 0 }^{ 2 }={ V }_{ x }^{ 2 }+{ V }_{ y }^{ 2 }##.....(4)

So ##{ V }_{ 0 }^{ 2 }={ V }_{ x }^{ 2 }+{ \left( \frac { 25 }{ 2{ V }_{ x } } \right) }^{ 2 }##

##V_{0}## will be minimum when ##V_{0} ^{2}## will be minimum.

So differentiating the equation wrt ##V_{x}## and equating to 0 I got that this function would have minimum value when ##V_{x}=\frac { 5 }{ \sqrt { 2 } } ##. So ##V_{y}=\frac { 5 }{ \sqrt { 2 } } ##

Putting this in equation(4) I got minimum value of ##v_{0}=5m/s##.

But ##5m/s## is not in the options provided. I am doubting my solution. Am I right?

 

[PeroK]

Homework Statement

A frog sits on the end of a long board of length L=5m . The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board . What is the minimum take-off speed (in s.i. Units) i.e.,relative to ground 'v' that allows the frog to do the trick ? The board and the frog have equal masses.

Homework Equations

The Attempt at a Solution

View attachment 81756
[/B]
As no external force act in the horizontal direction on the system so horizontal velocity of the frog is equal to the velocity of the plank (their directions are opposite). And also the position of the CoM of the system at final position and initial position should be same.

Now , the CoM of the plank will be displaced by length ##5/2##.

So ##t=\frac{5}{2V_{x}}##.....(1)

In this time period the from will jump and land on the plank.

So ##t=\frac{V_{y}}{5}##.....(2)

Using eq(1) and (2) we get

##V_{y}V_{x}=\frac{25}{2}##...(3)

Let the net velocity of the frog be ##v_{0}##

So ##{ V }_{ 0 }^{ 2 }={ V }_{ x }^{ 2 }+{ V }_{ y }^{ 2 }##.....(4)

So ##{ V }_{ 0 }^{ 2 }={ V }_{ x }^{ 2 }+{ \left( \frac { 25 }{ 2{ V }_{ x } } \right) }^{ 2 }##

##V_{0}## will be minimum when ##V_{0} ^{2}## will be minimum.

So differentiating the equation wrt ##V_{x}## and equating to 0 I got that this function would have minimum value when ##V_{x}=\frac { 5 }{ \sqrt { 2 } } ##. So ##V_{y}=\frac { 5 }{ \sqrt { 2 } } ##

Putting this in equation(4) I got minimum value of ##v_{0}=5m/s##.

But ##5m/s## is not in the options provided. I am doubting my solution. Am I right?

Your logic that the plank moves 5/2 is sound. But, you could also see this as the frog and plank must have equal and opposite velocities in the x direction. So, the frog must jump 5/2 and the plank move 5/2.

This statement is wrong:

"In this time period the from will jump and land on the plank.

So ##t=\frac{V_{y}}{5}##.....(2)"

I'm not sure how you got that. Given gravity is involved.

Anyway, the problem is essentially to find the minimum speed for the frog to jump 5/2. Does that sound familiar?

 

[Delta2]

he just took g=10, and the time to go up equals the time to get down equal to [itex]V_y/g[/itex] so total time [itex]t=2/10 V_y[/itex]. But yes the problem is essentially the minimum initial speed for the frog to have a horizontal displacement 5/2 when it lands.

 

[Satvik Pandey]

"In this time period the from will jump and land on the plank.

So ##t=\frac{V_{y}}{5}##.....(2)"

I'm not sure how you got that. Given gravity is involved.

Let the vertical velocity of the frog be ##V_{y}## So time required to reach the highest position is ##V_{y}/g##

So total time of flight would be ##2V_{y}/g##

Your logic that the plank moves 5/2 is sound. But, you could also see this as the frog and plank must have equal and opposite velocities in the x direction. So, the frog must jump 5/2 and the plank move 5/2.

How you proved that frog only jumps 5/2?

"In this time period the from will jump and land on the plank.

Why is this wrong?

 

[PeroK]

Let the vertical velocity of the frog be ##V_{y}## So time required to reach the highest position is ##V_{y}/g##

So total time of flight would be ##2V_{y}/g##
How you proved that frog only jumps 5/2?Why is this wrong?

I thought you had divided by the length of the plank(!), but I see from what Delta said that you are using ##g = 10 ms^{-2}##.

Your solution is almost correct, you just made a sily mistake taking square roots at the end.

 

[Satvik Pandey]

I thought you had divided by the length of the plank(!), but I see from what Delta said that you are using ##g = 10 ms^{-2}##.

Your solution is almost correct, you just made a sily mistake taking square roots at the end.

Thank you PeroK . Could you please point the mistake. I am unable to find it.

 

[PeroK]

Thank you PeroK . Could you please point the mistake. I am unable to find it.

Apologies, my mistake. I agree with your answer.

 

[Delta2]

what options your answer key gives?
maybe there is slight difference due to g, maybe there is a different value given for g like g=9.81?

 

[Satvik Pandey]

Apologies, my mistake. I agree with your answer.

Thank you!

what options your answer key gives?
maybe there is slight difference due to g, maybe there is a different value given for g like g=9.81?

Other potions were 1,2,3 and 4. But the answer of this question is ##2m/s##. I think it is wrong.

Thank you PeroK and Delta².