Physics 1 Practice Exam — Free

Resolve the initial velocity. Horizontal: $v_{0x} = 15.0\cos 30.0^\circ = 12.99$ m/s, which stays constant because no horizontal force acts. Vertical, taking up as positive: $v_{0y} = 15.0\sin 30.0^\circ = 7.50$ m/s.

Find the flight time from the vertical motion. The ball lands 20.0 m below its launch point, so $\Delta y = -20.0$ m:

The negative root is discarded because it is before the throw. The horizontal distance is then

As a check, the vertical velocity at impact is $v_y = 7.50 - 9.81(2.924) = -21.2$ m/s, and $v_y^2 = 449 = 7.50^2 + 2(9.81)(20.0)$, as energy conservation requires.

The tempting answer 19.9 m is the range on level ground, $v_0^2\sin 60^\circ/g$, which ignores the 20 m drop. 26.2 m uses the free-fall time from 20 m ($\sqrt{2h/g} = 2.02$ s) and so ignores the upward launch; 43.9 m uses the full 15.0 m/s as the horizontal speed.

$x = 38.0\,\text{m}$

Draw a free-body diagram for each block and apply Newton's second law to each separately. Because the string is inextensible and the pulley is massless and frictionless, both blocks share the same acceleration magnitude $a$ and the same tension $T$.

Block on the table (mass $m_1 = 4.00$ kg), taking the direction toward the pulley as positive: the only horizontal force is the tension, since the table is frictionless. The normal force and the weight cancel vertically.

Hanging block (mass $m_2 = 2.00$ kg), taking downward as positive:

Add the two equations to eliminate $T$; the tension is internal to the two-block system and the hanging weight is the only external force along the string:

The tension is less than the hanging block's weight (19.6 N), as it must be: if $T$ equalled the weight, the hanging block could not accelerate downward. The second equation confirms the numbers: $m_2 g - T = 19.62 - 13.08 = 6.54$ N, which equals $m_2a = 2.00(3.27)$.

$a = 3.27\,\text{m/s}^2$, $T = 13.1\,\text{N}$

Forces on the cylinder: weight $Mg$ downward, normal force $N$ perpendicular to the ramp, and static friction $f$ up the ramp at the contact point. Friction provides the torque that spins the cylinder, so it must point up the slope.

(a) Newton's second law for translation along the ramp (down-slope positive) and for rotation about the centre:

Rolling without slipping links the two motions, $a = \alpha R$, so the torque equation becomes $f = \tfrac{1}{2}Ma$. Substituting:

(b) $f = \tfrac{1}{2}Ma = \tfrac{1}{2}(5.00)(2.764) = 6.91$ N, directed up the ramp.

(c) The acceleration is constant, so $v^2 = 2a\,d = 2(2.764)(3.00) = 16.58$, giving $v = 4.07$ m/s. Energy conservation gives the same result: static friction does no work in rolling, so $Mgd\sin\theta = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 = \tfrac{3}{4}Mv^2$, and $v = \sqrt{\tfrac{4}{3}(9.81)(3.00)\sin 25.0^\circ} = 4.07$ m/s.

(d) Static friction must supply 6.91 N without exceeding its maximum $\mu_s N$, where $N = Mg\cos\theta = 5.00(9.81)\cos 25.0^\circ = 44.5$ N:

The cylinder accelerates at only two-thirds of the frictionless-sliding value $g\sin\theta$ because a third of the gravitational work goes into rotational kinetic energy.

(a) $a = 2.76\,\text{m/s}^2$; (b) $f = 6.91\,\text{N}$ up the ramp; (c) $v = 4.07\,\text{m/s}$; (d) $\mu_{s,min} = 0.155$