Analysis of Two Blocks Sliding Down Frictionless Inclined Planes

Consider the scenario where two blocks slide down separate, frictionless inclined planes. Block 1 slides down a plane inclined at 50 degrees, while Block 2 slides down a 40-degree plane. Additionally, Block 2 has an initial velocity of 2 m/s. Both blocks start simultaneously from the same initial height, and the question is: at what time will they be at the same height again?

Understanding the Problem and Key Variables

First, let's identify the key variables:

Block 1: Inclined plane at 50 degrees, starting from rest Block 2: Inclined plane at 40 degrees, with an initial velocity of 2 m/s Va: Velocity of Block 1: 0 gsin(50)t gsin(50)t Vb: Velocity of Block 2: 2 gsin(40)t H: Initial height from which both blocks start h: Height at which both blocks meet again g: Acceleration due to gravity (9.81 m/s2)

To find the time ( t ) when the blocks are at the same height again, we will use the conservation of energy and the kinematic equations.

Solving the Problem Using Energy Conservation

Using the conservation of energy, we can write the following for the two blocks:

For Block 1:
[ mgH mgh frac{1}{2}mVa^2 ]

For Block 2:
[ mgH - frac{1}{2}mV_o^2 mgh frac{1}{2}mV_b^2 ]

Where ( V_o 2 , text{m/s} ) and ( a g sin(50) ), ( b g sin(40) ).

Step 1: Express Height in Terms of Time

For Block 1, the height ( h ) can be expressed as:

[ h H - frac{1}{2}g(Va)^2 ]

For Block 2, the height ( h ) can be expressed as:

[ h H - frac{1}{2}g(Vb)^2 - V_o^2 ]

Step 2: Equate the Expressions for Height

By equating the two expressions for ( h ), we get:

[ frac{1}{2}g(Va)^2 frac{1}{2}g(Vb)^2 V_o^2 ]

Substitute the velocities:

[ frac{1}{2}g(g sin(50)t)^2 frac{1}{2}g(g sin(40)t 2)^2 2^2 ]

Step 3: Solve the Quadratic Equation

Divide through by (frac{1}{2}g), simplify, and solve for ( t ):

[ g sin(50)t^2 g sin(40)t^2 2 cdot 2g sin(40)t 4 ]

Combine like terms:

[ g sin(50)t^2 - g sin(40)t^2 - 4 cdot g sin(40)t - 4 0 ]

Using the values ( sin(50) approx 0.766 ) and ( sin(40) approx 0.643 ):

[ 10 cdot 0.766t^2 - 10 cdot 0.643t^2 - 4 cdot 10 cdot 0.643t - 4 0 ]

Simplify further:

[ 7.66t^2 - 6.43t^2 - 25.72t - 4 0 ][ 1.23t^2 - 25.72t - 4 0 ]

Step 4: Solve the Quadratic Equation for ( t )

Use the quadratic formula ( t frac{-b pm sqrt{b^2 - 4ac}}{2a} ):

[ t frac{25.72 pm sqrt{25.72^2 4 cdot 1.23 cdot 4}}{2 cdot 1.23} ]

Calculate the discriminant:

[ t frac{25.72 pm sqrt{661.5184 19.68}}{2.46} ][ t frac{25.72 pm sqrt{681.1984}}{2.46} ][ t frac{25.72 pm 26.10}{2.46} ]

Discard the negative root since time cannot be negative:

[ t frac{51.82}{2.46} approx 21.06 , text{seconds} ]

However, the problem requires a precise solution, and we need to check the context of the problem to ensure accuracy:

Final Solution and Verification

After revisiting the problem, it was identified that the vertical distance comparison is required, not the distance down the ramp. Using the vertical distance expressions, the correct time ( t ) can be found using:

[ t frac{4 sin(40) g}{g sin(50) - sin(40)} ]

With ( g 9.8 , text{m/s}^2 ), ( sin(40) approx 0.643 ), and ( sin(50) approx 0.766 ):

[ t frac{4 cdot 0.643 cdot 9.8}{9.8 cdot 0.766 - 0.643} approx 2.8 , text{seconds} ]

This solution is more accurate and precise. Thus, the blocks will be at the same height again after approximately 2.8 seconds.