PHY force, motion, mechanical

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2011-12-06 1:29 am

1. a ball is projected up a frictionless incline at an initial speed of 19.6m/s. the angle of the incline is 30.
a) what is the component of acceleraton due to gravity directed down the incline?
b) determine the time taken for the ball to return to its starting position
c) determine the distance the ball moves up the slope.

2. a jet is traveling at a speed of 600km/h at an angle of 30degree to the ground.
a) what is the horizontal component of its speed?
b) what is the vertical component of its speed?
c) hoe high would it climb in 30s at this speed and rate of climb?

3. a jet flying 500km/h N in level flight encounters a wind of 150km/h blowing from the SW
a) if no corrections are made, what will be the airspeed of the jet relative to the grounf?
b) what will be the new direction of the jet's velocity?
c) what correction to flight path would the pilot need to make to maintain the original flight path?

4.A net horizontal force of 5N acts on a 10kg mass initially at rest on a smooth horizatal surface. How long will it take for the mass to acquire a speed of 10m/s.

5. an air rifle is fired vertically at a block of wood supported between 2 tables as shown, The air rifle slug has a mass of 0.5g and its speed is 150m/s. If the block of wood rises vertically, what vertical height above the table will it rise to?

6. a 200
圖片參考：http://imgcld.yimg.com/8/n/HA07982681/o/701112050053813873407800.jpg

6. a 200g ball moving at a speed of 40m/s has an elastic collision with a stationary ball. After the collision the 200g ball moves in its original direction with a speed of 10m/s. what is the mass of the stationary ball?

回答 (1)

Prof. Physics

2011-12-06 3:01 am

✔ 最佳答案

1.a. Acceleration directed down the incline, a = gsin@ = (9.8)sin30* = 4.9 ms^-2

b. By s = ut + 1/2 at^2

0 = (19.6)t + 1/2 (-4.9)t^2

t = 0 (rejected) or 8s

c. At top of the slope, v = 0

By v^2 = u^2 + 2as

0 = (19.6)^2 + 2(-4.9)s

Distance, s = 39.2 m

2.a. Horizontal speed = vcos@ = 600cos30* = 520 ms^-1

Vertical speed = vsin@ = 600sin30* = 300 ms^-1

c. Height gained = vtsin@ = 300(30) = 9000 m

3.a. Consider the wind, the north-south velocity component, v = 150sin45* to the North

So, the airplane is 500 + 150sin45* km/h to the North (in this direction)

The west-east component, w = 150cos45* to the east

Airspeed relative to the ground, W = sqrt[(v + V)^2 + w^2]

= sqrt[(500 + 150sin45*)^2 + (150cos45*)^2]

= 615 ms^-1

New Angle to the North, @ = tan^-1 [w/(V + v)] = tan^-1 [(150cos45*) / (150sin45* + 500)] = 9.93*

So, the direction is N9.93*E

c. Assuming the jet fly with the same speed relative to the wind.

To fly in the original flight path, the airplane must have a westward velocity component

So, 500sin@ = 150sin45*

@ = 12.25*

The direction of the airplane should head is N12.25*W

4. By F = ma

5 = 10a

a = 0.5 ms^-2

By v = u + at

10 = 0 + 0.5t

Time required, t = 20 s

5. Assume the rifle stick to the block after the impact

By conservation of momentum

mu = (M + m)v

(0.0005)(150) = (0.05 + 0.0005)v

Rising Velocity of the block, v = 1.485 ms^-1

After that, energy is conserved.

By K.E. lost = G.P.E. gained

1/2 (M + m)v^2 = (M + m)gh

Height, h = v^2 / 2g = (1.485)^2 / 2(9.8) = 0.113 m

6. By conservation of momentum

mu = mv1 + Mv

(0.2)(40) = (0.2)(10) + Mv

Mv = 6 ... (1)

As the collision is elastic, energy is conserved,

By energy conservation

Initial K.E. = Final K.E.

1/2 mu^2 = 1/2 mv1^2 + 1/2 Mv^2

(0.2)(40)^2 = (0.2)(10)^2 + (6)(6/M)

36/M = 300

Mass, M = 0.12 kg = 120 g

參考: Prof. Physics

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