Answer to Question #95179 in Classical Mechanics for Jake

Question

2. Two marbles are launched at t = 0 in the experiment illustrated in the figure below. Marble 1 is launched horizontally with a speed of 4.20 m/s from a height h = 0.950 m. Marble 2 is launched from ground level with a speed of 5.94 m/s at an angle above the horizontal. (a) Where would the marbles collide in the absence of gravity? Give the x and y coordinates of the collision point. (b) Where do the marbles collide given that gravity produces a downward acceleration of g = 9.81 m/s2? Give the x and y coordinates.

Answer 1

Question (a)

Components of the marble 2

• Horizontal component

$cos(45^{0})=\frac{V_{2o}}{V_{2ox}} \\ V_{2ox}=V_{2o}cos(45^{0})\\ V_{2ox}=5.94\frac{m}{s}cos(45^{0}) \\V_{2ox}=4.20\frac{m}{s}$

• Vertical component.

$sin(45^{0})=\frac{V_{2o}}{V_{2oy}} \\ V_{2oy}=V_{2o}sin(45^{0})\\ V_{2oy}=5.94\frac{m}{s}sin(45^{0}) \\V_{2oy}=4.20\frac{m}{s}$

When there is no gravity.

Marble 1 will follow a horizontal path.

Marble 2 moves from the ground to the height of marble 1 in parallel, therefore it is necessary to calculate the time it takes to travel the height h =0.950m

The time it takes for marble 2 to collide with marble 1 is given by:

$v_{oy}=\frac{Y}{t} \\t=\frac{Y}{V_{oy}}$

Where:

• Vertical velocity component $V_{oy}=4.20\frac{m}{s}$
• Distance $Y=h--->h=0.950m$
• time $t$

Numerically evaluating

$t=\frac{0.950m}{4.20\frac{m}{s}}\\t=0.226s$

The X coordinate where they collide is given by:

$X=X_{o}+V_{1ox}*t$

Where

• Marble Horizontal Velocity 1 $V_{1ox}=4.20\frac{m}{s}$
• Initial position $X_{o}=0m$
• time $t=0.226s$

Evaluating numerically. $X=0m+4.20\frac{m}{s}*0.226s=0.949m$

Finally both marbles collide at a height of h = 0.950m and at a horizontal distance of x =0.949m

The coordinate (x, y) where they collide is equal to $\boxed{(0.949,0.950)m}$

Question (b)

Knowing that the horizontal displacement is the same, it is necessary to calculate when the vertical positions of both marbles collide in the presence of gravity.

The vertical position for marble 1 is given by:

$Y_{1}=Y_{o1}+V_{1oy}*t+\frac{1}{2}a_{y}t^{2}$

Where

• Initial position $Y_{1o}=h --->h=0.950m$
• Initial vertical velocity. $V_{1oy}=0\frac{m}{s}$
• Gravity acceleration a$a_{y}=-9.8\frac{m}{s^{2}}$
• time $t$

Expression.

$Y_{1}=h+\frac{1}{2}a_{y}t^{2}$

The vertical position for marble 1 is given by:

$Y_{2}=Y_{2o}+V_{2oy}*t+\frac{1}{2}a_{y}t^{2}$

Where

• Initial position $Y_{2o}=0m$
• Initial vertical velocity. $V_{2oy}=4.20\frac{m}{s}$
• Gravity acceleration $a_{y}=-9.8\frac{m}{s^{2}}$
• time $t$

Expression

$Y_{2}=V_{2oy}*t+\frac{1}{2}a_{y}t^{2}$

Both marbles collide when

$Y_{1}=Y_{2} \\ h+\frac{1}{2}a_{y}t^{2}=V_{2oy}*t+\frac{1}{2}a_{y}t^{2}$

Simplifying

$h=V_{2oy}*t$

solving for t

$t=\frac{h}{V_{2oy}}$

Numerically evaluating

$t=\frac{0.950m}{4.20\frac{m}{s}}\\t=0.226s$

The vertical position where they are found is calculated using any of the position equations of marble 1 or marble 2

The Y position is

$Y_{1}=0.950m-\frac{1}{2}9.81\frac{m}{s^{2}}(0.226s)^{2}=0.700m$

$Y_{2}=4.20\frac{m}{s}*0.226s-\frac{1}{2}9.81\frac{m}{s^{2}}(0.226s)^{2} =0.700m$

The horizontal distance is $X_{1}=X_{0}+V_{ox}*t$

Where:

• Initial position $X_{o}=0m$
• The horizontal velocity $V_{ox}=4.20\frac{m}{s}$
• Time $t=0.226s$

Numerically evaluating

$X=0m+4.20\frac{m}{s}*0.226s=0.949m$

In the presence of gravity the marbles are in the coordinates

$\boxed{(0.949,0.700)m}$