Dynamics Practice Exam — Free

Use normal–tangential components. The tangential acceleration is constant, so the constant-acceleration relation $v^2 = v_0^2 + 2a_t s$ applies along the path:

$$a_t = \frac{v^2 - v_0^2}{2s} = \frac{20^2 - 10^2}{2(75)} = \frac{300}{150} = 2.00\,\text{m/s}^2$$

The normal (centripetal) component depends only on the instantaneous speed and the radius of curvature:

$$a_n = \frac{v^2}{\rho} = \frac{15^2}{200} = 1.125\,\text{m/s}^2$$

The two components are perpendicular, so the magnitude is their vector sum:

$$a = \sqrt{a_t^2 + a_n^2} = \sqrt{2.00^2 + 1.125^2} = \sqrt{5.266} = 2.29\,\text{m/s}^2$$

The acceleration vector points $\tan^{-1}(1.125/2.00) = 29.4^\circ$ inward from the tangent.

The tempting answer 3.13 m/s² adds the two components as scalars; they are at right angles, so they must be combined with Pythagoras. 2.00 m/s² ignores the normal component, which exists whenever the path curves, and 1.13 m/s² ignores the speed-up.

$a = 2.29\,\text{m/s}^2$

Apply the principle of work and energy between the release point (1) and the point 4.0 m down the slope (2): $T_1 + \sum U_{1\to 2} = T_2$, with $T_1 = 0$ because the block starts from rest.

Forces on the block: weight $W = 5.0(9.81) = 49.05$ N, the normal force $N = W\cos 30^\circ = 42.48$ N (there is no acceleration perpendicular to the plane), and kinetic friction $F_f = \mu_k N = 0.20(42.48) = 8.496$ N acting up the slope. The normal force does no work because it is perpendicular to the motion.

Work of gravity (positive, since the block drops $4.0\sin 30^\circ = 2.0$ m):

$$U_g = W\,s\sin 30^\circ = 49.05(4.0)(0.5) = 98.1\,\text{J}$$

Work of friction (negative, since it opposes the motion):

$$U_f = -F_f\,s = -8.496(4.0) = -33.98\,\text{J}$$

Work–energy:

$$0 + 98.1 - 33.98 = \tfrac{1}{2}(5.0)v_2^2 \;\Rightarrow\; v_2^2 = \frac{2(64.12)}{5.0} = 25.65$$ $$v_2 = 5.06\,\text{m/s}$$

Without friction the speed would be $\sqrt{2(9.81)(2.0)} = 6.26$ m/s; friction removes about a third of the kinetic energy, which is consistent with a friction force that is about a third of the down-slope weight component.

$v = 5.06\,\text{m/s}$

Take rightward as positive, so $(v_A)_1 = +6.0$ m/s and $(v_B)_1 = -2.0$ m/s.

Conservation of linear momentum for the two-sphere system (no external impulse acts along the line of impact):

$$3.0(6.0) + 5.0(-2.0) = 3.0\,v_A + 5.0\,v_B \;\Rightarrow\; 3v_A + 5v_B = 8.0$$

Coefficient of restitution, written as separation velocity over approach velocity so that the signs take care of themselves:

$$e = \frac{v_B - v_A}{(v_A)_1 - (v_B)_1} \;\Rightarrow\; v_B - v_A = 0.60\,[6.0 - (-2.0)] = 4.8$$

Substitute $v_B = v_A + 4.8$ into the momentum equation:

$$3v_A + 5(v_A + 4.8) = 8.0 \;\Rightarrow\; 8v_A = -16 \;\Rightarrow\; v_A = -2.0\,\text{m/s},\quad v_B = 2.8\,\text{m/s}$$

Sphere A rebounds to the left at 2.0 m/s and sphere B moves to the right at 2.8 m/s. The spheres separate ($v_B > v_A$), as they must after an impact.

Kinetic energy before and after:

$$T_1 = \tfrac{1}{2}(3.0)(6.0)^2 + \tfrac{1}{2}(5.0)(2.0)^2 = 54 + 10 = 64\,\text{J}$$ $$T_2 = \tfrac{1}{2}(3.0)(2.0)^2 + \tfrac{1}{2}(5.0)(2.8)^2 = 6.0 + 19.6 = 25.6\,\text{J}$$

The loss $T_1 - T_2 = 38.4$ J is dissipated as heat, sound and permanent deformation. A loss is expected whenever $e < 1$; it is large here (60% of the initial energy) because the spheres approach each other quickly while their centre of mass moves at only $8.0/8.0 = 1.0$ m/s.

(a) $v_A = 2.0$ m/s to the left, $v_B = 2.8$ m/s to the right; (b) 38.4 J of kinetic energy is lost