Physics Chapter: Impulse and Momentum

<p>The force is not constant during the collision. (A)</p> Signup and view all the answers

<p>The weight of the ball is negligible compared to the force exerted by the bat. (A)</p> Signup and view all the answers

<p>Newton-second (N  s) (A)</p> Signup and view all the answers

<p>The impulse of the engines is equal to the change in momentum of the shuttle. (A)</p> Signup and view all the answers

<p>$\Sigma F \Delta t = mv_f - mv_i$ (C)</p> Signup and view all the answers

<p>PE = mgh (B)</p> Signup and view all the answers

<p>watt-second (W-s) (A)</p> Signup and view all the answers

<p>v = √(2gh) (B)</p> Signup and view all the answers

<p>xcm = (m1x1 + m2x2)/(m1 + m2) (D)</p> Signup and view all the answers

<p>Wgravity = mg(ho - hf) (B)</p> Signup and view all the answers

<p>W = Fs (A)</p> Signup and view all the answers

<p>xcm = (x1 + x2)/2 (A)</p> Signup and view all the answers

<p>Work done against Gravity (C)</p> Signup and view all the answers

<p>200 N (B)</p> Signup and view all the answers

<p>20 kg (C)</p> Signup and view all the answers

<p>1000 N (C)</p> Signup and view all the answers

<p>Both B and C are correct. (D)</p> Signup and view all the answers

<p>One-fifth of the way (C)</p> Signup and view all the answers

<p>0 N (D)</p> Signup and view all the answers

<p>Because the bolt is pulling down on the board. (A)</p> Signup and view all the answers

<p>Because the problem specifically instructs to ignore the weight of the board. (B)</p> Signup and view all the answers

<p>They are equal and opposite. (A)</p> Signup and view all the answers

<p>2.50 m (A)</p> Signup and view all the answers

<p>F2 would increase. (A)</p> Signup and view all the answers

<p>Static equilibrium. (A)</p> Signup and view all the answers

<p>The required mass would increase. (A)</p> Signup and view all the answers

<p>20,000 kg*m/s (B)</p> Signup and view all the answers

<p>0 kg*m/s (C)</p> Signup and view all the answers

<p>0 N (D)</p> Signup and view all the answers

<p>4 m/s (D)</p> Signup and view all the answers

<p>8 m/s (B)</p> Signup and view all the answers

<p>It is doubled (D)</p> Signup and view all the answers

<p>4000 N (D)</p> Signup and view all the answers

<p>Momentum is always conserved in a closed system (B)</p> Signup and view all the answers

<p>$\frac{\sum_{v=1}^{N} m_v r_v}{M}$ (C)</p> Signup and view all the answers

<p>$\frac{\sum_{v=1}^{N} m_v \dot{r}_v}{M}$ (C)</p> Signup and view all the answers

<p>$\frac{\sum_{v=1}^{N} m_v x_v}{M}$ (C)</p> Signup and view all the answers

<p>6.5 (C)</p> Signup and view all the answers

<p>Represents the average location of the system's mass. (D)</p> Signup and view all the answers

<p>The acceleration of the center of mass must be zero. (D)</p> Signup and view all the answers

<p>The force applied to the object multiplied by the lever arm. (B)</p> Signup and view all the answers

<p>The mass of the object. (A)</p> Signup and view all the answers

The product of mass and velocity (p = mv) of an object.

The impulse of a net force equals the change in momentum of an object.

The product of average force and time (ΣF ∆t) during a collision.

Kilogram meter per second (kg.m/s) is the unit measure for momentum.

The net force equals mass times acceleration (ΣF = ma).

Difference between final and initial momentum (mvf - mvo).

Force exerted by the bat that affects the ball's momentum.

Applied forces during events like the shuttle launch increase momentum.

The quantity of motion of a moving body, measured as a product of its mass and velocity.

The change in momentum per unit time; calculated as F = Δp/Δt.

The energy an object has due to its motion, calculated as KE = 1/2 mv².

A collision in which both momentum and kinetic energy are conserved.

Using kinetic energy and momentum to solve for mass; m = P/v.

The speed of an object in a specific direction.

The force exerted by rain when it strikes a surface, calculated as F = -m * vo.

Energy an object has due to its height above ground, calculated as PE = mgh.

The energy transferred when a force moves an object, defined as W = (F cosθ)s.

The speed of an object at the start of its motion, calculated using vo = - √(2g(ho - hf)).

The speed of an object at the end of its motion, calculated from initial speed and work done.

The average location of a system's mass, calculated as xcm = (m1x1 + m2x2) / (m1 + m2).

A collection of particles whose motion is analyzed together, often using center of mass coordinates.

The vertical distance an object is from a reference point, often ground level in physics.

An interaction that changes the motion of an object, often calculated in Newtons (N).

The average position of all particles' mass in a system.

A vector representing an object's position from a reference point.

The total mass of all particles in the system.

The speed and direction of the center of mass over time.

A measure of how much a force causes an object to rotate.

The product of force and lever arm distance, τ = Fℓ.

Distance from the axis of rotation to the line of action of the force.

Calculating changes in the position to find velocity and acceleration.

In equilibrium, sum of forces and torques is zero.

The total of all external forces acting on an object, must equal zero in equilibrium (ΣF = 0).

The total of all torques about a point must be zero (Στ = 0) in equilibrium.

The support force exerted upward by the fulcrum on a board in equilibrium.

Torque is calculated as force times distance from pivot (τ = F × d).

The downward force (F1) acting to prevent counterclockwise rotation on a board.

Weight of an object creates a clockwise torque about the fulcrum.

The length of the board affects torque calculation by changing lever arms.

The sum of torques equals zero when in equilibrium: ∑τ = F λ1 - W λW = 0.

The total vertical forces must balance: ΣFy = -F1 + F2 - W = 0.

Weight (W) is calculated as W = mg, where g = 10 m/s².

To find mass, use F = mg and rearrange: m = F/g.

For a bridge, support forces depend on weight distribution and position of loads.