Can You Solve These Challenging Physics Problems?

In summary, the conversation includes five physics problems that require help. The first problem involves two canoeists paddling upstream and downstream with given velocities, and the questions ask for the speed of the water and the relative speeds of the canoes. The second problem involves two connected crates on an incline, and the questions ask for the acceleration of one crate and the tension in the string. The third problem involves a ball being thrown with a bounce and without a bounce, and the questions ask for the angle at which the ball should be thrown to achieve the same distance and the ratio of times for the two throws. The fourth problem involves a dart gun being fired from a horizontal position and a student sliding down an incline while holding the gun, with
  • #1
tristan4040
1
0
A Few Tough Physics Problems! I'm stuck, Please Help!

1. Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One paddles directly upstream (and moves upstream), whereas the other paddles directly downstream. With downstream as the positive direction, an observer on shore determines the velocities of the two canoes to be -1.2 m/s and +2.9 m/s, respectively.

(a) What is the speed of the water relative to shore?
m/s

(b) What is the speed of each canoe relative to the water? canoe going upstream m/s
canoe going downstream m/s





2. Two packing crates of masses m1 = 10.0 kg and m2 = 6.50 kg are connected by a light string that passes over a frictionless pulley as in Figure P4.26. The 6.50 kg crate lies on a smooth incline of angle 39.0°. Find the acceleration of the 6.50 kg crate.

a) m/s2 (up the incline)

b) Find the tension in the string.
N





3. When baseball outfielders throw the ball, they usually allow it to take one bounce on the theory that the ball arrives sooner this way. Suppose that after the bounce the ball rebounds at the same angle θ as it had when released but loses half its speed.

(a) Assuming the ball is always thrown with the same initial speed, at what angle θ should the ball be thrown in order to go the same distance D with one bounce as one thrown upward at ϕ = 48.2° with no bounce?
°

(b) Determine the ratio of the times for the one-bounce and no-bounce throws.
t1b / t0b =





4. A dart gun is fired while being held horizontally at a height of 0.90 m above ground level and while it is at rest relative to the ground. The dart from the gun travels a horizontal distance of 4.00 m. A college student holds the same gun in a horizontal position while sliding down a 45.0° incline at a constant speed of 2.00 m/s. How far will the dart travel if the student fires the gun when it is 1.00 m above the ground?

m





5. A 4.9 g bullet leaves the muzzle of a rifle with a speed of 332 m/s. What force (assumed constant) is exerted on the bullet while it is traveling down the 0.85 m long barrel of the rifle?
N
 
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  • #2


Hi tristan4040,

You must show what work you've attempted for these problems before you can receive help. We will not do your homework for you. You must at least be able to show what equations/concepts you might need to use.
 
  • #3


1. (a) To find the speed of the water relative to shore, we can use the equation Vwater = Vcanoe + Vobserver. The canoe going upstream has a velocity of -1.2 m/s and the canoe going downstream has a velocity of +2.9 m/s. Plugging these values into the equation, we get Vwater = -1.2 m/s + 2.9 m/s = 1.7 m/s. Therefore, the speed of the water relative to shore is 1.7 m/s.

(b) The speed of each canoe relative to the water can be found by subtracting the speed of the water from the observed velocities. For the canoe going upstream, Vcanoe = Vobserver - Vwater = -1.2 m/s - 1.7 m/s = -2.9 m/s. For the canoe going downstream, Vcanoe = Vobserver - Vwater = 2.9 m/s - 1.7 m/s = 1.2 m/s. Therefore, the canoe going upstream has a speed of -2.9 m/s and the canoe going downstream has a speed of 1.2 m/s relative to the water.

2. (a) To find the acceleration of the 6.50 kg crate, we can use the equation Fnet = ma. The only force acting on the crate is the tension in the string, which is pulling the crate up the incline. The force of gravity is balanced by the normal force from the incline. Therefore, Fnet = T - mg sin θ = ma. Solving for a, we get a = (T - mg sin θ) / m = (T - 6.50 kg * 9.8 m/s^2 * sin 39.0°) / 6.50 kg = 2.86 m/s^2.

(b) To find the tension in the string, we can use the same equation Fnet = ma. The only force acting on the 6.50 kg crate is the tension in the string. Therefore, Fnet = T = ma = 6.50 kg * 2.86 m/s^2 = 18.6 N.

3. (a) To find the angle θ, we can use the equation D = (v0^2 / g) * sin 2ϕ, where D is
 

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