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To solve the problem, we must follow several steps to find both the distance and angle at which the two cars skid before coming to a stop after their inelastic collision. The steps include calculating the final velocity of the system using momentum conservation, the resulting kinetic energy, and how friction acts to stop the vehicles.

So, here’s a step-by-step solution detailed in the table format:

Step Derivation/Formula Reasoning
1 m_1 = 1000 \text{ kg}, v_{1x} = 20 \text{ m/s}, m_2 = 2000 \text{ kg}, v_{2y} = 15 \text{ m/s} Set initial conditions with masses of the cars and their velocities. The subscript 1 refers to the eastbound car, and 2 refers to the northbound car.
2 \vec{p}_{\text{initial}} = m_1 v_{1x} \hat{i} + m_2 v_{2y} \hat{j} Calculate the initial momentum vectorally considering east as \hat{i} direction and north as \hat{j}.
3 \vec{p}_{\text{final}} = \vec{p}_{\text{initial}} = (20000 \hat{i} + 30000 \hat{j}) \, \text{kg m/s} In an inelastic collision, total momentum is conserved. Calculate final momentum (which is the same as initial since there are no external forces in the horizontal plane).
4 \vec{v}_{\text{final}} = \frac{\vec{p}_{\text{final}}}{m_1 + m_2} = \frac{20000 \hat{i} + 30000 \hat{j}}{3000} = (6.67 \hat{i} + 10 \hat{j}) \, \text{m/s} Calculate the final velocity vector of the system combining both masses (the mass after collision is sum of both masses).
5 v_{\text{final}} = \sqrt{6.67^2 + 10^2} \approx 12.08 \, \text{m/s} Calculate the magnitude of final velocity using the Pythagorean theorem.
6 \theta = \tan^{-1} \left( \frac{10}{6.67} \right) Calculate the angle of the final velocity vector using tangent inverse. Angle is measured from the east axis (positive x-axis).
7 F = \mu_k m g Calculate the friction force acting, which opposes the motion. \mu_k is the coefficient of kinetic friction and g is acceleration due to gravity (approximately 9.8 \, \text{m/s}^2).
8 \Delta KE = 0.5 (m_1 + m_2) v_{\text{final}}^2 Calculate the kinetic energy just after the collision.
9 \text{Work done by friction, } W = F \cdot d = \Delta KE Friction does work to bring the cars to a stop, equal to the change in kinetic energy (which is all converted into heat and other forms).
10 d = \frac{\Delta KE}{F} = \frac{0.5 \cdot 3000 \cdot 12.08^2}{0.9 \cdot 3000 \cdot 9.8} \approx 8.19 \, \text{m} Calculate the skidding distance. Plugging in all the known values will give the distance over which the cars skid.

Summary:
– Distance skidded: 8.19 meters.
– Angle from east direction: \tan^{-1}(1.5) \approx 56.31° north of east.

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8.19 m, 56.3°

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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