1. Why will a sheet of paper fall slower than one that is crumpled into a ball?
2. Gravitational force on the surface of the moon is only 1/6 as strong as gravitational force on the earth. What is the weight in newton’s of a 10 kg object on the moon and on the earth?
3. A ball is thrown vertically upwards with a velocity of 49 m/s.
Calculate
(i) The maximum height to which it rises,
(ii) The total time it takes to return to the surface of the earth.
4. A stone is released from the top of a tower of height 19.6 m. Calculate its final velocity just before touching the ground.
5. A stone is thrown vertically upward with an initial velocity of 40 m/s. Taking g = 10 m/s2, find the maximum height reached by the stone. What is the net displacement and the total distance covered by the stone.
Answers
Answer:
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Explanation:
1.A sheet of paper falls slower then the one crumpled into a ball because the surface area of the ball is much lower than that of the sheet. Thus the upward force exerted by air on the sheet is much as compared to the one exerted on the ball. Hence the sheet falls slower.
2.check-circle
Text Solution
Solution :
Mass of object,m=10kgm=10kg
This is the same on earth and moon .
weight of the object on earth, We=mge=10×9.8=98NWe=mge=10×9.8=98N
weight of the object on moon, Wm=mgm=10×9.86=166.3N
Gravitational force on the surface of the moon is only 1/6 as gravitational force on the earth. What is the weight in newtons of a 10kg object on the moon and on the earth? This is the same on earth and moon . weight of the object on moon, Wm=mgm=10×9.86=166.3N.
3.Given Initial velocity of ball, u=49 m/s
Let the maximum height reached and time taken to reach that height be H and t respectively.
Assumption: g=9.8 m/s
2
holds true (maximum height reached is small compared to the radius of earth)
Velocity of the ball at maximum height is zero, v=0
v
2
−u
2
=2aH
0−(49)
2
=2×(−9.8)×H
⟹H=122.5 m
v=u+at
0=49−9.8t
⟹t=5 s
∴ Total time taken by ball to return to the surface, T=2t=10 s
4.Initial Velocity u=0
Fianl velocity v=?
Height, s=19.6m
By third equation of motion
v
2
=u
2
+2gs
v
2
=0+2×9.8×19.6
v
2
=384.16
⇒v=19.6m/s
5.Initial Velocity u=40
Fianl velocity v=0
Height, s=?
By third equation of motion
v
2
−u
2
=2gs
0−40
2
=−2×10×s
s=
20
160
⇒s=80m/s
Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m
Total Diaplacement =0, Since the initial and final point is same.
other solution :
Initial Velocity u=40
Fianl velocity v=0
Height, s=?
By third equation of motion
v
2
−u
2
=2gs
0−40
2
=−2×10×s
s=
20
160
⇒s=80m/s
Toatl distance travelled by stone = upward distance + downwars distance =2×s=160m
Total Diaplacement =0, Since the initial and final point is same.