Law and Collisions t Explorer Bailistic Pendulum previous 16 of 18 I net s Balis

Question

Law and Collisions t Explorer Bailistic Pendulum previous 16 of 18 I net s Balistic Pendulum .balsk dum an objed of mass "n ls fired with an initial speed to at a pendulum bob The bob has a mass f, which is suspended by a rod of length L and negligible mass After the collision, the pendulum and object stick together and swing to a maximum angular dglacernent as shown(Bart Find an expressionfo", nad speed of the fired eb d Expeess your answer in terms of some or all of the variables m, M. L and 8 and the acceleration due to grevity, , mints "" . (M + m)g1(1 cost) ncorrect Try Again; 3 attempts remaining Part An experiment is done to compare the initial speed of bulets fired from different handguns a 0u and a 44 callber. The guns are fired into a 10-kg pendulum bob of length L Assume hat the 90 mn bullet has a mass of 60 g and the 44-calber bullet has a mass of 12 the 0-mm bullet causes the pendulum to swing to a maximum angular dsplacement of 4 3" and he 44 calber bullet causes a dsplacement of 10 1 find the ratio of the initial speed of the mm bullet to the speed of the 44 -calber bulet (o/() Express your answe numerically Hints

Explanation / Answer

Part A = ((m+M)/m)(sqrt(2gL(1-cos())))

Part B = 0.852

How to do Part A:

Find an expression for v0, the initial speed of the fired object.

Express your answer in terms of some or all of the variables m, M, L, and theta and the acceleration due to gravity, g.

This is a purely algebraic manipulation. Solving this requires a couple of steps. The first thing to do is figure out a relationship between the initial and final velocities. Since momentum is conserved, we can use that as a starting point:

p0 = mv0

pf = (m+M)v

Since momentum is conserved, we can say that:

p0 = pf

Therefore:

mv0 = (m+M)v

and:

v0 = ((m+M)v / m)

So the only thing left now is figure out an expression for vf. We can do this by analyzing the relationship between kinetic and potential energy:

We know that once point after the bullet hits the ballistic pendulum, the bullet and pendulum mass will fuse and the combined mass will take on kinetic energy. As the mass moves upwards, gravity will slow it down and some of the kinetic energy will be converted to potential energy. When the mass is at it’s maximum height, potential energy will be maximum and kinetic energy will be zero (gravity will have slowed the velocity to zero and the height will be maximized). Vice-versa, when the pendulum is at the bottom (zero height), potential energy will be zero and kinetic energy will be maximum.

We also know that the formula for kinetic energy has velocity in it – but we need a way to remove velocity from our expression. Potential energy doesn’t have velocity in the formula, and we know that at some point, when the pendulum is at a certain height, kinetic energy and potential energy will be equal. So we can set the two energies equal to each other, solve for velocity (in the kinetic energy side), and plug the result into our expression above:

U = (m+M)gh

KE = 1/2(m+M)v2

(m+M)gh = 1/2mv2

v = sqrt(2gh)

The problem is, the problem doesn’t let us use ‘h’ in our answer. But we can find h in terms of the angle and the length of the pendulum, L:

h = L(1 – cos())

Think about it this way- when the pendulum is just hanging, it’s height (since the string is holding it) will be zero. As it moves up, will increase, and if the string isn’t taut, the difference between the new length and the original length will have to be the height. For example, if the length is 1 meter and is 60°, assuming the string isn’t being held taut (it’s slack), 1-cos(60°) = 0.5 and 0.5L = 0.5. Now in reality the string won’t be 100% slack, but the mathematical formula still holds true because the angle is what’s important, not the length of the sides of the ‘triangle’.

So since we have h in terms of L and , we can plug this into our expression for v:

v = sqrt(2gL(1-cos()))

And now plug this into the first expression we solved for:

v0 = ((m+M)v / m)

v0 = ((m+M)/m)(v)

v0 = ((m+M)/m)(sqrt(2gL(1-cos())))

How to do Part B:

Using the formula from Part A, just solve for each v0 separately and then for the ratio:

9mm:

v0,9 = ((m+M)/m)((m+M)/m)(sqrt(2gL(1-cos())))

v0,9 = ((0.006+10)/0.006)(sqrt(2(9.8)L(1-cos(4....

v0,9 = 1667.67(sqrt(19.6L(0.0028)))

v0,9 = 1667.67(sqrt(0.05488L))

v0,9 = 1667.67(0.2343)(sqrt(L))

v0,9 = 390.74sqrt(L)

.44 caliber:

v0,44 = ((m+M)/m)((m+M)/m)(sqrt(2gL(1-cos())))

v0,44 = ((0.012+10)/0.012)(sqrt(2(9.8)L(1-cos(10...

v0,44 = 834.33(sqrt(19.6L(0.0155)))

v0,44 = 834.33(sqrt(0.3038L))

v0,44 = 834.33(0.55)(sqrt(L))

v0,44 = 458.88sqrt(L)

v0,9 / v0,44 = 390.74sqrt(L) / 458.88sqrt(L)

v0,9 / v0,44 = 0.852

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