Sliding down a sliding inclined plane

Classical Mechanics Level 4

A box of mass 2 kg 2\text{ kg} slides without friction down an inclined plane of mass 5 kg 5\text{ kg} that is free to slide horizontally on a frictionless floor. The height of the inclined plane where the box initially stands is 5 5 meters, and the inclination of the plane is 3 0 ∘ 30^{\circ} .

What will be the magnitude of the velocity of the box (with respect to the stationary floor) just before it leaves the inclined plane?


Details and Assumptions:

  • Neglect the dimensions of the box.
  • Take the gravitational acceleration as g = 9.81 m / s 2 g = 9.81\text{ m / s}^2 .


The answer is 8.88.

Hosam Hajjir by Hosam Hajjir , 56, Canada

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2 solutions

Steven Chase
Sep 14, 2017

Here's my approach. There's probably an easier way.

Let x x be the horizontal position of the bottom of the ramp, and let α \alpha be the distance of the box up the ramp. The horizontal and vertical positions of the box are x b x_b and y b y_b . θ \theta is the ramp angle. The box position is:

x b = x + α c o s θ y b = α s i n θ x b ˙ = x ˙ + α ˙ c o s θ y b ˙ = α ˙ s i n θ x_b = x + \alpha \, cos\theta \\ y_b = \alpha \, sin\theta \\ \dot{x_b} = \dot{x} + \dot{\alpha} \, cos\theta \\ \dot{y_b} = \dot{\alpha} \, sin\theta

Box kinetic energy:

E b = 1 2 m v b 2 = 1 2 m ( x b ˙ 2 + y b ˙ 2 ) = 1 2 m ( x ˙ 2 + 2 x ˙ α ˙ c o s θ + α ˙ 2 ) E_b = \frac{1}{2} m v_b^2 = \frac{1}{2} m (\dot{x_b}^2 + \dot{y_b}^2) = \frac{1}{2}m (\dot{x}^2 + 2 \, \dot{x} \dot{\alpha} \, cos\theta + \dot{\alpha}^2)

Ramp kinetic energy:

E r = 1 2 ( 5 2 m ) x ˙ 2 = 5 4 m x ˙ 2 E_r = \frac{1}{2} (\frac{5}{2}m) \dot{x}^2 = \frac{5}{4} m \dot{x}^2

Box gravitational potential energy:

U b = m g α s i n θ U_b = mg \, \alpha \, sin\theta

System Lagrangian:

L = E b + E r − U b = 7 4 m x ˙ 2 + m x ˙ α ˙ c o s θ + 1 2 m α ˙ 2 − m g α s i n θ L = E_b + E_r - U_b = \frac{7}{4} m \dot{x}^2 + m \, \dot{x} \, \dot{\alpha} \, cos\theta + \frac{1}{2} m \, \dot{\alpha}^2 - mg \, \alpha \, sin\theta

Equations of Motion:

d d t ∂ L ∂ x ˙ = ∂ L ∂ x d d t ∂ L ∂ α ˙ = ∂ L ∂ α \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{x}}} = \frac{\partial{L}}{\partial{x}} \\ \frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\alpha}}} = \frac{\partial{L}}{\partial{\alpha}}

Evaluating these equations results in:

7 2 x ¨ + c o s θ α ¨ = 0 c o s θ x ¨ + α ¨ + g s i n θ = 0 \frac{7}{2} \ddot{x} + cos\theta \, \ddot{\alpha} = 0 \\ cos\theta \, \ddot{x} + \ddot{\alpha} + g \, sin\theta = 0

Solving the system and plugging in for θ \theta gives:

α ¨ = − 7 11 g x ¨ = 3 11 g \ddot{\alpha} = -\frac{7}{11} g \\ \ddot{x} = \frac{\sqrt{3}}{11} g

Remaining steps:
1) Numerically integrate to solve for α ¨ \ddot{\alpha} , x ¨ \ddot{x} , α ˙ \dot{\alpha} , x ˙ \dot{x} , α \alpha , and x x over time. Initialize the variables per the problem description.
2) When α \alpha becomes zero, evaluate the expression v b = x ˙ 2 + 2 x ˙ α ˙ c o s θ + α ˙ 2 v_b = \sqrt{\dot{x}^2 + 2 \, \dot{x} \dot{\alpha} \, cos\theta + \dot{\alpha}^2} to find the block velocity.
3) Using g = 9.8 g = 9.8 , the answer comes out to approximately 8.88 \boxed{8.88}


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The easier way is as follows:

  • Momentum is conserved only in the horizontal direction. (The normal force by floor on ramp is an external vertical force.) If v v is the velocity of the box and V V the velocity of the ramp at a given time, then m v cos ⁡ θ + M V = 0 , mv\cos \theta + MV = 0, so that V = − m cos ⁡ θ M v . V = -\frac{m\cos \theta}M v.

  • The total kinetic energy of the system is K = 1 2 m v 2 + 1 2 M V 2 = ( M + m cos ⁡ 2 θ 2 M ) m v 2 . K = \tfrac12 m v^2 + \tfrac 12 M V^2 = \left( \frac {M + m\cos^2\theta} {2 M} \right) m v^2.

  • The decrease in potential energy is Δ U = − m g h . \Delta U = -mgh.

  • Since mechanical energy is conserved, we find for the final velocity K = ( M + m cos ⁡ 2 θ 2 M ) m v 2 = m g h ; K = \left( \frac {M + m\cos^2\theta} {2 M} \right) m v^2 = mgh; v = 2 g h ⋅ M M + m cos ⁡ 2 θ = 2 ⋅ 9.81 ⋅ 5 ⋅ 5 5 + 2 ⋅ 3 4 ≈ 8.687 m / s . v = \sqrt{2gh\cdot \frac{M}{M + m\cos^2\theta}} = \sqrt{2\cdot 9.81\cdot 5 \cdot \frac{5}{5 + 2\cdot \tfrac34}} \approx \SI{8.687}{m/s}.

Note that the factor M M + m cos ⁡ 2 θ = 1 1.3 \sqrt{\frac{M}{M + m\cos^2\theta}} = \frac 1{\sqrt{1.3}} is the "correction" factor for the fact that the ramp slides backward. In the case M → ∞ M \to \infty or θ → 9 0 ∘ \theta \to 90^\circ , this factor reduces to 1.

Arjen Vreugdenhil - 3 years, 8 months ago

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The equation for the conservation of linear momentum in the horizontal direction should read

m ( v cos ⁡ θ + V ) + M V = 0 m (v \cos \theta + V) + M V = 0

where v v is the velocity of the box with respect to the incline, and V V is the velocity of the incline with respect to the stationary floor.

This makes V = − m v cos ⁡ θ M + m = − r v V = -\dfrac{ m v \cos \theta }{M + m} = - r v , where r = m cos ⁡ θ M + m r = \dfrac{ m \cos \theta}{M + m}

And the kinetic energy of the system is

K = 1 2 m ( ( v cos ⁡ θ + V ) 2 + ( v sin ⁡ θ ) 2 ) + 1 2 M V 2 K = \frac{1}{2} m ( (v \cos \theta + V)^2 + (v \sin \theta)^2 ) + \frac{1}{2} M V^2

which upon simplifying becomes,

K = 1 2 m ( v 2 + V 2 + 2 v V cos ⁡ θ ) + 1 2 M V 2 K = \frac{1}{2} m ( v^2 + V^2 + 2 v V \cos \theta) + \frac{1}{2} M V^2

Upon substituting for V V , we get

K = 1 2 m ( v 2 + r 2 v 2 − 2 r v 2 cos ⁡ θ ) + 1 2 M r 2 v 2 K = \frac{1}{2} m (v^2 + r^2 v^2 - 2 r v^2 \cos \theta) + \frac{1}{2} M r^2 v^2

so that

K = 1 2 ( m ( 1 + r 2 − 2 r cos ⁡ θ ) + M r 2 ) v 2 K = \frac{1}{2} ( m (1 + r^2 - 2 r \cos \theta) + M r^2 ) v^2

Equating this with the change in potential energy, results in the desired v v .

v final = 2 m g h m ( 1 + r 2 − 2 r cos ⁡ θ ) + M r 2 v_{\text{final}} = \large \sqrt{ \dfrac{ 2 m g h } { m (1 + r^2 - 2 r \cos \theta) + M r^2 } }

The numerical value of r = 2 cos ⁡ π 6 5 + 2 = 0.24743582965 r = \dfrac{ 2 \cos \dfrac{\pi}{6} }{5 + 2} = 0.24743582965

Plugging this into the above expression, gives

v final = 2 ( 2 ) ( 9.81 ) ( 5 ) 2 ( 1 + r 2 − 2 r cos ⁡ θ ) + 5 r 2 = − 11.173833 v_{\text{final}} = \large \sqrt{ \dfrac{ 2 (2) (9.81) (5) } { 2 (1 + r^2 - 2 r \cos \theta) + 5 r^2 } } = -11.173833

The corresponding V = − r v = − 0.24743582965 ( − 11.173833 ) = 2.76480665584 V = - r v = - 0.24743582965 (-11.173833 ) = 2.76480665584

Finally, the velocity of the box with respect to the floor is ( v cos ⁡ θ + V ) (v \cos \theta + V) in the horizontal direction and v sin ⁡ θ v \sin \theta in the vertical direction, so that the magnitude is

∣ v abs = ( v cos ⁡ θ + V ) 2 + ( v sin ⁡ θ ) 2 ) = v 2 + V 2 + 2 v V cos ⁡ θ | v_{\text{abs}} = \sqrt{ (v \cos \theta + V)^2 + (v \sin \theta)^2 )} = \sqrt{ v^2 + V^2 + 2 v V \cos \theta}

Pluggin in v = − 11.173833 v = -11.173833 and V = 2.76480665584 V = 2.76480665584 results in

∣ v abs ∣ = v 2 + V 2 + 2 v V cos ⁡ θ = 8.8876 m / s | v_{\text{abs}} | = \sqrt{ v^2 + V^2 + 2 v V \cos \theta} = 8.8876{m/s}

Hosam Hajjir - 3 years, 8 months ago

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Ah yes, you are right of course :)

Arjen Vreugdenhil - 3 years, 8 months ago

Interesting. It seems we have two different camps, both claiming slightly different answers.

Steven Chase - 3 years, 8 months ago

Thanks for a well-written solution.

Hosam Hajjir - 3 years, 8 months ago

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You're welcome, and thanks. I'm planning on a follow-up with a different geometry.

Steven Chase - 3 years, 8 months ago

Does the answer comes out to be 981 13 \boxed{\sqrt{\frac{981}{13}}} in appropriate units?

Akshay Yadav - 3 years, 8 months ago

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That comes out to about 8.687, which isn't sufficiently close to the numerically derived answer. However, it should be possible to get an exact answer. I just didn't bother.

Steven Chase - 3 years, 8 months ago

I'm getting the same answer too! I used momentum conservation in horizontal direction and work energy theorem on the system.

Akshat Sharda - 3 years, 8 months ago

I did this using the exact same method that you did, the lagrangian method, but I couldn’t carry out the last steps, like numerically integrating it.

Krishna Karthik - 2 years, 7 months ago

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You should look into numerical integration. It's very useful

Steven Chase - 2 years, 7 months ago
Ahmed Aljayashi
Mar 19, 2019

v = 2 g h 4 M 2 + 2 M m + m 2 4 M 2 + 5 M m + m 2 ≈ 8.887 m s v =\sqrt{2gh\frac{4M^{2}+2Mm+m^{2}}{4M^{2}+5Mm+m^{2}}}\approx 8.887\ \frac{m}{s}

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