### Problem StatementA careless driver driving at 25 m/s hit a tree while fumbling with his phone. Assume the driver was in contact with the airbag for 0.22 seconds, and his upper body has 37 kg in mass. How much average force (absolute value) does the airbag exert on the driver's upper body?**Ignore gravity since the force was horizontal.****Hint:** Use impulse and the change of momentum.### Solution ApproachTo solve the problem, we need to follow these steps:1. **Determine the initial and final velocities:** - Initial velocity (\(v_i\)) = 25 m/s - Final velocity (\(v_f\)) = 0 m/s (since the driver comes to a stop)2. **Calculate the change in momentum (\( \Delta p \)):** - The momentum (\( p \)) of an object is given by \( p = m \times v \). - Change in momentum (\( \Delta p \)) = \( m \times (v_f - v_i) \).3. **Use the impulse-momentum theorem:** - Impulse ( \( J \) ) = Change in momentum (\( \Delta p \)). - Impulse is also given by \( J = F \times t \), where \( F \) is the force and \( t \) is the time interval.4. **Solve for the average force (\( F \)):** - Rearrange the impulse formula to solve for \( F \): \( F \) = \( \frac{\Delta p}{t} \).Given:- Initial velocity, \( v_i \) = 25 m/s- Final velocity, \( v_f \) = 0 m/s- Mass, \( m \) = 37 kg- Time, \( t \) = 0.22 seconds### Detailed Calculation1. **Change in Momentum** \[ \Delta p = m \times (v_f - v_i) \] \[ \Delta p = 37 \, \text{kg} \times (0 - 25 \, \text{m/s}) \] \[ \Delta p = 37 \, \text{kg} \times -25 \, \text{m/s} \] \[ \

Newton's second law of motion describes the net force acting on an object to be a product of mass…

A careless driver driving at 25 m/s hit a tree while fumbling with his phone. Assume the driver was in contact with the air bag for 0.22 seconds, and his upper body has 37 kg in mass, then how much average force (absolute value) does the air bag exert on the driver's upper body? Ignore gravity since the force was horizontal. Hint: use impulse and the change of momentum.

A careless driver driving at 25 m/s hit a tree while fumbling with his phone. Assume the driver was in contact with the air bag for 0.22 seconds, and his upper body has 37 kg in mass, then how much average force (absolute value) does the air bag exert on the driver's upper body? Ignore gravity since the force was horizontal. Hint: use impulse and the change of momentum.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

A 20 kg object and a 30 kg object collide. The figure below is for the 20 kg object. Assume momentum is conserved and find the change in velocity of the 30 kg object over the entire time shown. am/sis) 30.0 25.0 20.0 15.0 10.0 5.0 2.0 4.0 6.0 t(s)
Jaycee and Jonah are skating on an ice rink. As Jaycee approaches Jonah from west, he pushed her back providing her an impulse described by: J = (-75.0N/8)t² + 1/3(25.00N /s² )t³ due west. At what instant (in seconds) does Jaycee achieve maximum change in momentum?
The force shown in the force-time diagramacts on a 1.3 kg object. Find the impulse of the force.Answer in units of N · s. Find the final velocity of the particle if it isinitially at rest.Answer in units of m/s. Find the final velocity of the particle if itis initially moving along the x axis with avelocity of −1 m/s.Answer in units of m/s
A particle with a mass of 1.90 kg is acted on by a force Fx acting in the x-direction. If the magnitude of the force varies in time as shown in the figure below, determine the following. (a) impulse of the force (in kg · m/s ) kg · m/s (b) final velocity of the particle (in m/s) if it is initially at rest m/s (c) Find the final velocity of the particle (in m/s) if it is initially moving along the x-axis with a velocity of −25.00 m/s. m/s
1. a) If there is an external force acting on a system, can you please tell whether the momentum would be changed? If the momentum is changed, can you link the change in momentum to another physical quantity? 1 b) Define Impulse. Mention two units of Impulse. Is it a vector or scaler? If it is a vector, what would be the direction of impulse? 1 c) A golf ball with a mass 46.0 g strikes a wall with an initial speed of 140 km/h along –x- direction and bounces back at the speed of 145 km/h flying in xy plane, which makes 30° with the horizontal. Calculate the impulse vector in terms of magnitude and direction. If the duration of the collision is 0.900 ms, determine the magnitude of the average force acting on the ball during the collision. 2+1
Problem 2. A force acts on a 25.0 kg object with the magnitude as a function of time as shown in figure below. (a) Determine the impulse imparted by the force on the object between 0 and 30 seconds. (b) Determine the final velocity of the object if it had an initial velocity of 20.0 m/s in the negative direction. (c) Find the average force between t = 0 and 30 seconds. F,(N) 1000- t (s) 30 0- 10 20
If the impulse acting on a moving 100 kg object is zero, what does that mean about the velocity of the object? Recall that impulse is: F*t = m*Δv where Δv = vf - vi vf > vi vf = vi vf < vi Not enough information to determine.
When analyzing collisions using the law of conservation of momentum, the same principles that allow us to analyze collisions in one dimension can also be used to analyze collisions in two dimensions. To do this, you must separately apply the law of conservation of momentum in the x- and y-directions. Use the following example to practice this process: Two football players are about to collide with one another. Player 1 has a mass of 98.1 kg and is moving with a velocity of 4.52 m/s, 21.5° east of north. Player 2 has a mass of 79.4 kg and is moving with a velocity of 5.15 m/s, 8.55° east of south. Find the x- and y-components of the momentum of each player. What is the sum of the momenta in each dimension? If, when the players collide, they hold onto each other and move as a single unit, what will be the velocity vector of that single unit? Suppose that the two players collide, and bounce off each other. If player 2 is sent backwards with a velocity of 1.77 m/s, 15.5° west of north,…
A 0.0600 kg ball is traveling horizontally at 28.5 m/s when it is hit by a badminton racquet. The ball then moves in the opposite direction, also horizontally, at a speed of 44.5 m/s. Take the initial velocity of the ball to be in the -x-direction. (a) What is the impulse delivered to the ball by the badminton racquet? (Give the x-component in Ns. Indicate the direction with the sign of your answer.) ÎN S (b) Some work is done on the system of the ball and some energy appears in the ball as an increase in internal energy during the collision between the ball and the racket. What is the quantity W - AEint for the ball? (Give your answer in J.)
A rocket sled with a mass of 2900 kg moves at 250 m/s on a set of rails. At a certain point, a scoop on the sled dips into a trough of water located between the tracks and scoops water into an empty tank on the sled. By applying the principle of conservation of linear momentum, determine the speed of the sled after 920 kg of water has been scooped up. Ignore any retarding force on the scoop.
A 50 kg car is moving at a speed of 72 km/h when the driver applies the brakes that generate an impulse force modeled as shown in the graph below. What is the speed, in km/h, of the car after the break application? Round to the first decimal. Take g to be 10 m/s². Force (kN) x6x40 20 40 60 80 100 120 140 160 Time (ms)
For the situation in the figure below, use momentum conservation to determine (a) the magnitude and (b) the direction of the final velocity of ball 1 after the collision. The angle = 42.0⁰. (a) Number 01 = 0.900 m/s m₂ = 0.150 kg (b) Number i 02=0.540 m/s m₂ = 0.260 kg Units 35.0 (b) (a) Top view of two balls colliding on a horizontal surface. (b) This part of the drawing shows the x and y components of the velocity of ball 1 after the collision. Units 197x 2 = 0.700 m/s