OMPT mechanics practice: lesson and 10 problems

Answer: 1.5 m/s²

Acceleration is the change in velocity divided by the time it took: $a=\frac{\Delta v}{\Delta t}=\frac{13.0-4.0}{6.0}=\frac{9.0}{6.0}=1.5{\text{m/s}}^2$. Both speeds are already in metres per second, so nothing needs converting, and the unit falls out of the division: metres per second, per second.

Answer: B — $31\text{m}$

From rest, $s=\frac{1}{2}a{t}^2$ with $a=g$. So $s=\frac{1}{2}(9.81)(2.5{)}^2=\frac{1}{2}(9.81)(6.25)=30.66\text{m}$, which rounds to $31\text{m}$. Sanity check: the stone finishes at $v=gt=24.5\text{m/s}$, so its average speed over the fall is about $12.3\text{m/s}$, and $12.3\times 2.5\approx 31\text{m}$. The two routes agree.

Answer: C — $3.0{\text{m/s}}^2$

Newton's second law uses the resultant force, not the applied one. Horizontally, $F_{\text{res}}=18-6.0=12\text{N}$. Then $a=\frac{F_{\text{res}}}{m}=\frac{12}{4.0}=3.0{\text{m/s}}^2$. The vertical forces, weight and normal force, cancel and never enter the horizontal equation.

Answer: 50 N

Tilt the axes so one runs along the ramp. The weight $mg=15\times 9.81=147.15\text{N}$ splits into a component down the slope, $mg\sin \theta$, and one pressing into the slope, $mg\cos \theta$. Equilibrium along the ramp means the held force matches the first: $F=mg\sin 20^{\circ }=147.15\times 0.342=50.3\text{N}$, so $50\text{N}$ to two significant figures.

Answer: Step 2

Step 1 picks exactly the right equation: it is the one with no $t$ in it. The error is in Step 2, where $72\text{km/h}$ is substituted into a formula whose other quantities are in metres and seconds. Convert first: $72\text{km/h}=\frac{72000}{3600}=20\text{m/s}$. Then $a=\frac{0-20^2}{2\times 40}=-5.0{\text{m/s}}^2$, a deceleration of $5.0{\text{m/s}}^2$. Step 3 only repeats Step 2's number, so it is a consequence, not the mistake. The size of the error is its own warning: $64.8{\text{m/s}}^2$ is over six times gravity, which no road car produces.

Answer: 16 m

Split the motion into two independent one-dimensional problems. Vertically the ball starts with zero velocity and falls $20\text{m}$: $20=\frac{1}{2}(9.81)t^2$, so $t^2=4.077$ and $t=2.02\text{s}$. Horizontally there is no force, so the speed stays $8.0\text{m/s}$ and $x=8.0\times 2.02=16.2\text{m}$, or $16\text{m}$ to two significant figures.

Answer: C — $150\text{J}$

Work counts only the part of the force that lies along the displacement: $W=Fd\cos \theta$, where $\theta$ is the angle between them. Here $W=25\times 12\times \cos 60^{\circ }=300\times 0.5=150\text{J}$. The vertical component of the tension does no work at all, because the sledge does not rise.

Answer: 5.9 m/s

No friction means the gravitational potential energy all becomes kinetic energy: $mgh=\frac{1}{2}m{v}^2$. The mass cancels from both sides, so $v=\sqrt{2gh}=\sqrt{2\times 9.81\times 1.8}=\sqrt{35.3}=5.94\text{m/s}$, which is $5.9\text{m/s}$ to two significant figures. The $0.40\text{kg}$ is deliberately unused information.

Answer: 4800 N

The work-energy theorem says the work done by the brakes equals the kinetic energy removed. That energy is $\frac{1}{2}m{v}^2=\frac{1}{2}(1200)(20{)}^2=240000\text{J}$. Spread over $50\text{m}$, the force is $F=\frac{240000}{50}=4800\text{N}$. Cross-check through kinematics: $a=\frac{0-400}{2\times 50}=-4.0{\text{m/s}}^2$, and $F=ma=1200\times 4.0=4800\text{N}$. Two independent routes, one answer.

Answer: Step 1

Step 1 confuses mass with weight. Mass is $6.0\text{kg}$; weight is the force gravity exerts on it, $mg=6.0\times 9.81=58.9\text{N}$. The reasoning about constant speed is right, and Steps 2 and 3 carry the wrong number forward faithfully. With the correct force, $W=58.9\times 2.0=118\text{J}$. The same answer comes from the potential energy gained: $mgh=6.0\times 9.81\times 2.0=118\text{J}$.