Classical-Mechanics · Unit 12 · Video 7 · Interactive Practice

The Snag and the Slide: One Landing, Two Kinds of Physics

IKey Formulas

Formula	Name	What you need
$m_p v_{p,i} = (m_p + m_s)\,v_{p,1}$	Momentum across the snag	Both masses and the landing speed
$f_k = \mu_k N_{g,s} = \mu_k m_s g$	Kinetic friction on the sandbag	$\mu_k$ and the sandbag mass alone
$a_x = -\dfrac{F_{g,p} + \mu_k m_s g}{m_p + m_s}$	Newton II for the joined pair	Both retarding forces, combined mass
$x_p(t_f) = \dfrac{(m_p + m_s)\,v_{p,1}^{\,2}}{2\left(F_{g,p} + \mu_k m_s g\right)} = \dfrac{m_p^{\,2} v_{p,i}^{\,2}}{2(m_p + m_s)\left(F_{g,p} + \mu_k m_s g\right)}$	Stopping distance	Only the given quantities and $g$

Key Insight: Stopping distance is kinetic energy divided by total retarding force — and exactly one number, $v_{p,1}$ , crosses from the momentum stage into the force stage.

IIStage 1 — The Snag

What common speed carries exactly the momentum the plane brought into the snag?

Common speed after the snag (m/s) 22.0

💡 The snag is treated as instantaneous, so friction and the brakes deliver negligible impulse over it — that alone is what licenses $p_{x,i} = p_{x,1}$ .

IIIWhose Weight Reaches the Runway

Only what actually rests on the sandbag can set its normal force — what does getting that wrong cost?

Weight pressing through the sandbag Coefficient of kinetic friction 0.40

IVStage 2 — The Slide

Constant acceleration turns the post-snag speed into a distance: kinetic energy over total retarding force.

Time since the snag (s) 0.0

VQuiz Questions

Problem 1 · Speed Just After the Hook Catches

Given: a $1500\ \text{kg}$ plane rolling at $30\ \text{m/s}$ snags a $300\ \text{kg}$ sandbag at rest and drags it along — find the common speed $v_{p,1}$ immediately after the snag.

✅ Correct! The system gained

20\%

more mass and kept

83\%

of its speed, with the momentum

45\,000\ \text{kg}\cdot\text{m/s}

unchanged.

❌ The ratio is upside down. Adding mass at fixed momentum must lower the speed, so the factor is

m_p/(m_p+m_s)

, a number less than

1

, not its reciprocal.

❌ Not quite. Momentum, not speed, is what survives the snag — the same

45\,000\ \text{kg}\cdot\text{m/s}

now has to be carried by

1800\ \text{kg}

❌ Wrong mass on top. You used

m_s/(m_p+m_s)

; the momentum came in with the plane, so the numerator is

m_p

Show solution

The snag is instantaneous, so horizontal momentum is the same immediately before and after:

m_p v_{p,i} = (m_p + m_s)\,v_{p,1}

v_{p,1} = \frac{m_p}{m_p + m_s}\,v_{p,i} = \frac{1500}{1800}\,(30\ \text{m/s}) = 25\ \text{m/s}

Check: before, $p = (1500)(30) = 45\,000\ \text{kg}\cdot\text{m/s}$ ; after, $p = (1800)(25) = 45\,000\ \text{kg}\cdot\text{m/s}$ .

Problem 2 · Which Mass Sets the Friction

Given: a $900\ \text{kg}$ plane drags a $150\ \text{kg}$ sandbag along the runway by a horizontal cable, with $\mu_k = 0.5$ and $g = 9.8\ \text{m/s}^2$ — find the kinetic friction force $f_k$ on the sandbag.

✅ Correct! Only the sandbag's own weight presses on the runway through the sandbag, so

N_{g,s} = m_s g = 1470\ \text{N}

and

f_k = \mu_k N_{g,s} = 735\ \text{N}

❌ That is the classic trap. You used the total mass

1050\ \text{kg}

; the plane's weight rides on the plane's own wheels and never reaches the ground through the bag.

❌ That is the normal force, not the friction.

1470\ \text{N} = m_s g = N_{g,s}

— you still owe it a factor of

\mu_k = 0.5

❌ Wrong body.

900\ \text{kg}

is the plane; the vertical equation you need is written for the sandbag alone.

Show solution

Step 1 — vertical equation for the sandbag alone. The bag has no vertical acceleration, and the cable pulls horizontally, so it contributes nothing vertically:

N_{g,s} - m_s g = 0 \quad\Longrightarrow\quad N_{g,s} = (150\ \text{kg})(9.8\ \text{m/s}^2) = 1470\ \text{N}

Step 2 — kinetic friction.

f_k = \mu_k N_{g,s} = (0.5)(1470\ \text{N}) = 735\ \text{N}

Using the total mass instead would give $f_k = (0.5)(1050)(9.8) = 5145\ \text{N}$ , seven times too large here — and it would shrink the stopping distance by a comparable factor.

Problem 3 · Both Stages End to End

Given: an $800\ \text{kg}$ plane touches down at $36\ \text{m/s}$ and immediately snags a $100\ \text{kg}$ sandbag with $\mu_k = 0.5$ ; the brakes add a constant $1310\ \text{N}$ of retarding force and $g = 9.8\ \text{m/s}^2$ — find the post-snag speed and the distance travelled from the snag point to rest.

Speed immediately after the snag?

Distance from the snag point to rest?

✅ Correct!

v_{p,1} = 32\ \text{m/s}

feeds a constant

a_x = -2.0\ \text{m/s}^2

, and

x = v_{p,1}^{\,2}/(2|a_x|) = 256\ \text{m}

❌ The ratio is inverted.

v_{p,1} = m_p v_{p,i}/(m_p+m_s)

— the combined mass belongs in the denominator, so the speed must drop.

❌ Check stage one. Momentum is conserved across the snag:

(800)(36) = (900)\,v_{p,1}

❌ Wrong speed into stage two.

324\ \text{m}

uses the touchdown speed

36\ \text{m/s}

; the slide starts at the post-snag speed

32\ \text{m/s}

❌ The friction used the wrong mass. With

N = m_s g

the friction is

490\ \text{N}

, not

(0.5)(900)(9.8) = 4410\ \text{N}

❌ A force is missing.

9.4 \times 10^2\ \text{m}

comes from friction alone — the brakes contribute another

1310\ \text{N}

backward.

❌ Check stage two. Add both retarding forces, divide by the combined mass

900\ \text{kg}

, then use

x = v_{p,1}^{\,2}/(2|a_x|)

Show solution

Stage 1 — momentum across the snag:

v_{p,1} = \frac{m_p v_{p,i}}{m_p + m_s} = \frac{(800)(36)}{900} = 32\ \text{m/s}

Stage 2 — forces on the joined pair. The sandbag alone presses on the runway:

f_k = \mu_k m_s g = (0.5)(100)(9.8) = 490\ \text{N}

F_{g,p} + f_k = 1310 + 490 = 1800\ \text{N}

a_x = -\frac{1800\ \text{N}}{900\ \text{kg}} = -2.0\ \text{m/s}^2

Stage 2 — kinematics. Starting at $32\ \text{m/s}$ and stopping:

x = -\frac{v_{p,1}^{\,2}}{2 a_x} = \frac{(32\ \text{m/s})^2}{2(2.0\ \text{m/s}^2)} = 256\ \text{m}

Energy reading: $x = \tfrac{1}{2}(900)(32)^2 / 1800 = 460\,800/1800 = 256\ \text{m}$ — kinetic energy after the snag divided by the total retarding force.

Problem 4 · Scaling the Landing Speed

Given: the closed form $x_p(t_f) = \dfrac{m_p^{\,2} v_{p,i}^{\,2}}{2(m_p+m_s)\left(F_{g,p} + \mu_k m_s g\right)}$ , a second plane is identical in every way but touches down at $2v_{p,i}$ — find the factors by which the post-snag speed and the stopping distance change.

Post-snag speed $v_{p,1}$ is multiplied by?

Stopping distance $x_p(t_f)$ is multiplied by?

✅ Correct!

v_{p,1} \propto v_{p,i}

but

x_p \propto v_{p,i}^{\,2}

, so the original

382\ \text{m}

becomes about

1.5\ \text{km}

— while the deceleration never changes.

❌ Check stage one.

v_{p,1} = m_p v_{p,i}/(m_p+m_s)

is strictly proportional to

v_{p,i}

; the mass factor is untouched.

❌ Only the speed scales linearly. The distance carries

v_{p,i}^{\,2}

because it is kinetic energy divided by a force that does not depend on speed.

❌ Too steep.

v_{p,i}

appears squared in the numerator and nowhere else —

2^2 = 4

, not

2^3

❌ Look at the exponent. Every other symbol in

x_p(t_f)

is unchanged, so the factor is exactly

(2)^2

from

v_{p,i}^{\,2}

Show solution

Stage 1. The masses are unchanged, so

v_{p,1}' = \frac{m_p (2 v_{p,i})}{m_p + m_s} = 2 v_{p,1}

Stage 2. The acceleration $a_x = -(F_{g,p} + \mu_k m_s g)/(m_p+m_s)$ contains no speed at all, so it is identical for both planes. The distance is

x_p = \frac{v_{p,1}^{\,2}}{2|a_x|} \quad\Longrightarrow\quad x_p' = \frac{(2 v_{p,1})^2}{2|a_x|} = 4\,x_p

The same conclusion is visible in the closed form: $v_{p,i}$ appears only as $v_{p,i}^{\,2}$ , so doubling it multiplies the distance by $4$ . The video's $382\ \text{m}$ would become $1528\ \text{m}$ , and the stopping time would double from $21.4\ \text{s}$ to $42.8\ \text{s}$ .

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