Laws of Motion Test

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Laws of Motion Test - 10

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Weekly Quiz Competition

Question 1
1 / -0

A truck starts from rest and accelerates uniformly at 2.0 ms^-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What is the magnitude of acceleration (in ms^-2) of the stone at t = 11s? (Neglect air resistance.)

A
14 ms^-2

B
8 ms^-2

C
10 ms^-2

D
12 ms^-2

Solution

When the stone is dropped from the truck , the horizontal force acting on it become zero after stone is released. Stone continue to move under the influence of gravity only. So that acceleration of the stone is equal to the gravitational acceleration g.

a = g = 10ms^-2 acting vertically downward.
Question 2
1 / -0

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s⁻¹.What is the trajectory of the bob if the string is cut when the bob is at one of its extreme positions

A
bob will go down in a parabolic path

B
bob will fall vertically downwards

C
bob will go downwards at an angle

D
bob will go upwards

Solution

At the extreme position of the oscillation, the speed of the bob is zero. So the bob is momentarily at rest. if the string is cut, the bob will fall vertically downwards.
Question 3
1 / -0

A bob of mass 0.1 kg hung from the ceiling of a room by a string 2 m long is set into oscillation. The speed of the bob at its mean position is 1 m s⁻¹.What is the trajectory of the bob if the string is cut when the bob is at its mean position?

A
bob will go down in a parabolic path

B
bob will fall vertically downwards

C
bob will go upwards

D
bob will fall vertically downwards

Solution

At the mean position, the bob has a horizontal velocity. so when the string is cut, it will fall along a parabolic path under the effect of gravity.
Question 4
1 / -0

A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 5 m s⁻² what would be the reading on the scale?

A
130 kg

B
105 kg

C
150 kg

D
120 kg

Solution

When the lift moves upward with acceleration = 5 ms^-2 the net force acting upward

R−mg=ma

R=mg+ma

R=m(g+a)

R=70(10+5)

R=1050N

R = 1050 N

(We experience weight due to reaction)

therefore Apparent weight = =1050/g = 1050/10 = 105 kg
Question 5
1 / -0

A man of mass 70 kg stands on a weighing scale in a lift which is freely falling under gravity. What would be the reading on the scale?

A
125 kg

B
95 kg Hz

C
0 kg

D
115 kg

Solution

When the lift fall freely under gravity, a = g

Therefore Apparent weight, (We experience weight due to reaction)

R = m(g − a) = m(g − g) = 0

This is the condition for weightlessness.
Question 6
1 / -0

A man of mass 70 kg stands on a weighing scale in a lift which is moving upwards with a uniform speed of 10 m s⁻¹. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

A
105 kg

B
105 kg

C
35 kg

D
0 kg

Solution

When the lift falls freely under gravity, a=g

Therefore Apparent weight,

R = m(g − a) = m(g − g) = 0

This is the condition of weightlessness.

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Re-Attempt Weekly Quiz Competition

Laws of Motion Test - 10

Result

Laws of Motion Test - 10

Score

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Rank

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SHARING IS CARING

Question 1

14 ms-2

8 ms-2

10 ms-2

12 ms-2

Solution

Question 2

bob will go down in a parabolic path

bob will fall vertically downwards

bob will go downwards at an angle

bob will go upwards

Solution

Question 3

bob will go down in a parabolic path

bob will fall vertically downwards

bob will go upwards

bob will fall vertically downwards

Solution

Question 4

130 kg

105 kg

150 kg

120 kg

Solution

Question 5

125 kg

95 kg Hz

0 kg

115 kg

Solution

Question 6

105 kg

105 kg

35 kg

0 kg

Solution

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14 ms^-2

8 ms^-2

10 ms^-2

12 ms^-2