Two graduate students are playing a game in which they try to hit a small box on

Question

Two graduate students are playing a game in which they try to hit a small box on the floor with a marble fired from a spring loaded gun that is mounted on a table. The target box (center of the box) is horizontal distance D = 2.20m from the edge of the table (see figure below). Bobby compresses the spring 1.10 cm, but the center of the marble falls 27.0 cm short of the center of the box. Consider this as the “first shot”. How far should Rhoda compress the spring to score a direct hit? Consider the direct hit as the “second shot”. Assume that neither the spring nor the ball encounters friction in the gun. I will break the problem in small parts to walk you through the solution. Pay attention to different steps based on the specific questions I ask you below.

1. Let D1 be the horizontal distance to the landing point in the first shot and v1 be the velocity of the marble at the moment it leaves the nozzle in the first shot. In other words, v1 is the launch velocity in the first shot. Assume the launch velocity is horizontal. Also assume h is height of the table from the ground. Using your ideas of projectile motion, derive a relation that relates only D1, v1, h and g. As usual g is the acceleration due to gravity and it is positive. We are not going to use the numerical value of g yet.

2. Let, D2 be the horizontal distance to the landing during the “second shot” and v2 be the launch velocity at this shot. Once again, assume at the time of launch the vertical component of the velocity is zero. Derive an equation that relates only D2, v2, h and g. You could directly use ideas from the previous part.

3. Show that v2/v1 = D2/D1.

4. Let l1 be the compression of the spring during the “first shot”. Derive a relation only in terms of m, v1, k and l1. Let k be the spring constant.

5. Let l2 be the compression of the spring during the “second shot”. Derive a relation only in terms of m, v2, k and l2. Hint: think about the result derived above.

Explanation / Answer

1)

along vertical

y = (1/2)*g*t^2

h = (1/2)*g*t^2

time taken t = sqrt(2h/g)

along horizontal

x = vx*t

D1 = v1*sqrt(2h/g) <<<---------answer

=++++++++++++++++++===========

2)

along vertical

y = (1/2)*g*t^2

h = (1/2)*g*t^2

time taken t = sqrt(2h/g)

along horizontal

x = vx*t

D2 = v2*sqrt(2h/g) <<<---------answer

--------------------------

3)

from the above 2 eq

v2/v1 = D2/D1

-----------------------

elastic potential energy of spring = KE of the marble

(1/2)*k*l1^2 = (1/2)*m*v1^2

k*l1^2 = m*v1^2

---------------------------

elastic potential energy of spring = KE of the marble

(1/2)*k*l2^2 = (1/2)*m*v2^2

k*l2^2 = m*v2^2

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