What Forces Act on a Car During a Turn and Collision?

[jac5038]

1) A 1150 kg car moving at 5.7 m/s is initially traveling north in the positive y direction. After completing a 90.° right-hand turn to the positive x direction in 4.8 s, the inattentive operator drives into a tree, which stops the car in 450 ms.
(a) In unit-vector notation, what is the impulse on the car during the turn? x-component? y-component?
(b) In unit-vector notation, what is the impulse on the car during the collision? x-component? y-component?
(c) What is the magnitude of the average force that acts on the car during the turn?
(d) What is the magnitude of the average force that acts on the car during the collision?
(e) What is the angle between the average force in (c) and the positive x direction?

2)A pellet gun fires ten 2.0 g pellets per second with a speed of 560 m/s . The pellets are stopped by a rigid wall.
(a) What is the momentum of each pellet?
(b) What is the kinetic energy of each pellet?
(c) What is the average force exerted by the stream of pellets on the wall?
(d) If each pellet is in contact with the wall for 0.50ms, what is the average force exerted on the wall by each pellet during contact? Why is this average force so different from the average force calculated in (c)?

3)A railroad flatcar of weight 7840 N can roll without friction along a straight horizontal track. Initially, a man of weight 682 N is standing on the car, which is moving to the right with speed 25 m/s; see the figure. What is the change in velocity of the car if the man runs to the left so that his speed relative to the car is 4.5 m/s?

4)A 6.05 g bullet is fired horizontally at two blocks resting on a smooth tabletop. The bullet passes through the first block, with mass 1.20 kg, and embeds itself in the second, with mass 1.80 kg. Speeds of 0.390 m/s and 0.80 m/s, respectively, are thereby imparted to the blocks. Neglecting the mass removed from the first block by the bullet, find the speed of the bullet immediately after it emerges from the first block. Find the bullet's original speed.

5)The last stage of a rocket is traveling at a speed of 8450 m/s. This last stage is made up of two parts that are clamped together, namely, a rocket case with a mass of 310.00 kg and a payload capsule with a mass of 150.00 kg. When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of 910 m/s. What are the speeds of the two parts after they have separated? Assume that all velocities are along the same line. What is the speed of the payload? What is the speed of the rocket case? Find the total kinetic energy of the two parts before and after they separate; account for any difference. The kinetic energy before they separate. The kinetic energy after they separate.

6)A shell is fired from a gun with a muzzle velocity of 28 m/s, at an angle of 60° with the horizontal. At the top of the trajectory, the shell explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically. How far from the gun does the other fragment land, assuming that the terrain is level and that the air drag is negligible?

7) A cart of mass M1 = 4.0 kg and initial speed = 8.0 m/s collides head on with a second cart of mass M2 = 2.0 kg at rest. Assuming that the collision is elastic, find the speed of M2 after the collision.

 

[Tom McCurdy]

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[thepatientmental]

1) (a) The impulse on the car during the turn can be written as J = mΔv, where m is the mass of the car and Δv is the change in velocity. Since the car is initially traveling north in the positive y direction and turns 90° to the positive x direction, the change in velocity can be written as Δv = 5.7 m/s + 5.7 m/s = 11.4 m/s in the x direction. Therefore, the impulse on the car during the turn is J = (1150 kg)(11.4 m/s) = 13110 kg·m/s. In unit-vector notation, this can be written as J = (13110 kg·m/s) i. The x-component of this impulse is Jx = 13110 kg·m/s and the y-component is Jy = 0 kg·m/s.

(b) The impulse on the car during the collision can be written as J = mΔv, where m is the mass of the car and Δv is the change in velocity. Since the car is stopped in 450 ms, the change in velocity can be written as Δv = 5.7 m/s in the negative x direction. Therefore, the impulse on the car during the collision is J = (1150 kg)(-5.7 m/s) = -6555 kg·m/s. In unit-vector notation, this can be written as J = (-6555 kg·m/s) i. The x-component of this impulse is Jx = -6555 kg·m/s and the y-component is Jy = 0 kg·m/s.

(c) The magnitude of the average force that acts on the car during the turn can be calculated using the equation FΔt = mΔv, where F is the average force, Δt is the time interval, m is the mass of the car, and Δv is the change in velocity. Therefore, the average force can be calculated as F = (1150 kg)(11.4 m/s)/(4.8 s) = 2712.5 N.

(d) The magnitude of the average force that acts on the car during the collision can be calculated using the equation FΔt = mΔv, where F is the average force, Δt is the time interval, m is the mass of the car, and