75 Collisions of Extended Bodies in Two Dimensions

Suppose the disk in Figure 2 has a mass of 50.0 g and an initial velocity of 30.0 m/s when it strikes the stick that is 1.20 m long and 2.00 kg. (a) What is the angular velocity of the two after the collision? (b) What is the kinetic energy before and after the collision? (c) What is the total linear momentum before and after the collision?

Strategy for (a)

We can answer the first question using conservation of angular momentum as noted. Because angular momentum is Iω, we can solve for angular velocity.

Solution for (a)

Conservation of angular momentum states

where primed quantities stand for conditions after the collision and both momenta are calculated relative to the pivot point. The initial angular momentum of the system of stick-disk is that of the disk just before it strikes the stick. That is,

where I is the moment of inertia of the disk and ω is its angular velocity around the pivot point. Now, I = mr² (taking the disk to be approximately a point mass) and ω = v/r, so that

[latex]L={{mr}^{2}}\frac{v}{r}={mvr}\\[/latex].

After the collision,

It is ω′ that we wish to find. Conservation of angular momentum gives

[latex]I'\omega'={mvr}\\[/latex].

Rearranging the equation yields

[latex]\omega' =\frac{mvr}{I'}\\[/latex],

where I′ is the moment of inertia of the stick and disk stuck together, which is the sum of their individual moments of inertia about the nail. Figure 3 gives the formula for a rod rotating around one end to be I=Mr²/3. Thus,

[latex]I'={mr}^{2}+\frac{{Mr}^{2}}{3}=\left(m+\frac{M}{3}\right){r}^{2}\\[/latex].

Entering known values in this equation yields,

[latex]I'=\left(0.0500\text{ kg}+0.667\text{ kg}\right){\left(1.20\text{ m}\right)}^{2}=1.032\text{ kg}\cdot{\text{m}}^{2}\\[/latex].

The value of I′ is now entered into the expression for ω′, which yields

[latex]\begin{array}{lll}\omega'&=& \frac{mvr}{I'}=\frac{\left(0.0500\text{ kg}\right)\left(30.0\text{ m/s}\right)\left(1.20\text{ m}\right)}{1.032\text{ kg}\cdot\text{m}^{2}}\\ & =& 1.744\text{ rad/s}\approx 1.74\text{ rad/s}\end{array}\\[/latex].

Strategy for (b)

The kinetic energy before the collision is the incoming disk’s translational kinetic energy, and after the collision, it is the rotational kinetic energy of the two stuck together.

Solution for (b)

First, we calculate the translational kinetic energy by entering given values for the mass and speed of the incoming disk.

[latex]\text{KE}=\frac{1}{2}{mv}^{2}=\left(0.500\right)\left(0.0500\text{ kg}\right){\left(30.0\text{ m/s}\right)}^{2}=22.5 \text{ J}\\[/latex]

After the collision, the rotational kinetic energy can be found because we now know the final angular velocity and the final moment of inertia. Thus, entering the values into the rotational kinetic energy equation gives

[latex]\begin{array}{lll}\text{KE'}& =& \frac{1}{2}I'{\omega'^{2}}=\left(0.5\right)\left(1.032\text{kg}\cdot\text{m}^{2}\right)\left(1.744\frac{\text{rad}}{\text{s}}\right)^{2}\\ & =& 1.57\text{ J}\end{array}\\[/latex].

Strategy for (c)

The linear momentum before the collision is that of the disk. After the collision, it is the sum of the disk’s momentum and that of the center of mass of the stick.

Solution of (c)

Before the collision, then, linear momentum is

[latex]p=\text{mv}=\left(0\text{.}\text{0500}\text{kg}\right)\left(\text{30}\text{.}0\text{m/s}\right)=1\text{.}\text{50}\text{kg}\cdot \text{m/s}\\[/latex].

After the collision, the disk and the stick’s center of mass move in the same direction. The total linear momentum is that of the disk moving at a new velocity v′ = rω′ plus that of the stick’s center of mass, which moves at half this speed because [latex]{v}_{\text{CM}}=\left(\frac{r}{2}\right)\omega' =\frac{v'}{2}\\[/latex]

Thus,

Gathering similar terms in the equation yields,

[latex]p'=\left(m+\frac{M}{2}\right)v'\\[/latex]

so that

[latex]p' =\left(m+\frac{M}{2}\right)r\omega'\\[/latex].

Substituting known values into the equation,

[latex]p' =\left(1\text{.}\text{050 kg}\right)\left(1\text{.}\text{20}\text{m}\right)\left(1\text{.}\text{744 rad/s}\right)=\text{2.20 kg}\cdot \text{m/s}\\[/latex].

Discussion

First note that the kinetic energy is less after the collision, as predicted, because the collision is inelastic. More surprising is that the momentum after the collision is actually greater than before the collision. This result can be understood if you consider how the nail affects the stick and vice versa. Apparently, the stick pushes backward on the nail when first struck by the disk. The nail’s reaction (consistent with Newton’s third law) is to push forward on the stick, imparting momentum to it in the same direction in which the disk was initially moving, thereby increasing the momentum of the system.