Motion In a Straight Line-Quiz-5

Physics Quiz-5

Dear Readers,

If an object changes its position with respect to its surroundings with time, then it is called in motion. It is a change in the position of an object over time. Motion in a straight line is nothing but linear motion. As the name suggests, it’s in a particular straight line, thus it can be said that it uses only one dimension.

Q1. If two balls of same density but different masses are dropped from a height of 100 m, then (neglect air resistance)

•  Both will come together on Earth
•  Lighter will come earlier on Earth
•  Heavier will come earlier on Earth
•  None of these

Solution

The only force acting on both will be gravity which will produce same acceleration g in both. Further, both the balls are dropped simultaneously from same height, hence both will come together on the ground.

Q2.A car accelerates from rest at a constant rate a for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

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Solution

Let the car accelerate at rate α for time t_1 then maximum velocity attained,

v=0+at_1=at_1

Now, the car decelerates at a rate β for time (t-t_1) and finally comes to rest. Then,

0=v-β(t-t_1 )⇒0=αt_1-βt+βt_1

⇒t_1=β/(α+β) t

∴v=αβ/(α+β) t

Q3. The average velocity of a body moving with uniform acceleration after travelling a distance of 3.06 m is 0.34 ms^(-1). If the change in velocity of the body is 0.18 ms^(-1) during this time, its uniform acceleration is:

•  0.01 ms^(-2)
•  0.02 ms^(-2)
•  0.03 ms^(-2)
•  0.04 ms^(-2)

Solution

Given v_av=(v+u)/2=0.34 and v-u=0.18

Solving these two equations, we get

u=0.25 ms^(-1),v=0.43 ms^(-1). Given s=3.06 m

Now use v^2-u^2=2as to find a=0.02 ms^(-2)

Q4. A wooden block is dropped from the top of a cliff 100 m high and simultaneously a bullet of mass 10 g is fired from the foot of the cliff upwards with a velocity of 100 ms^(-1). The bullet and wooden block will meet after a time

•  10 s
•  0.5 s
•  1 s
•  7 s
  

Solution

Relative acceleration of both will be zero w.r.t. each other

So, s_rel=u_rel t or 100=100t or t=1 s

Q5.A particle starts from rest. Its acceleration (a)  versus time (t) is as shown in the figure.  The maximum speed of the particle will be

•  110 m/s
•  55 m/s
•  550 m/s
•  660 m/s

Solution

Area under acceleration-time graph gives the change in velocity.  Hence,

                       v_max=1/2×10×11=55 ms^(-1)

Therefore, the correct option is (b).

Q6. 

The given graph shows the variation of velocity with displacement. Which one of the graph                given below correctly represents the variation of acceleration with displacement?

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Solution

The v-x equation from the given graph can be written as,

                  v=(-v_0/x_0 )  x+v_0                       …(i)

∴             a=dv/dt=(-v_0/x_0 )  dx/dt=(-v_0/x_0 )  v

Substituting v from Eq. (i), we get

          a=(-v_0/x_0 )[(-v_0/x_0 )  x+v_0 ]

         a=(v_0/x_0 )^2  x-(v_0^2)/x_0 

Thus, a-x graph is a straight line with positive slope and negative intercept.

Q7.

A particle moving in a straight line covers half the distance with speed of 3 m/s.

The other half of the distance is covered in two equal time intervals with speed

of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during the motion is

•  4.0 m/s
•  5.0 m/s
•  5.5 m/s
•  4.8 m/s

Solution

If t_1 and 2t_2 are the time taken by particle to cover first and second half distance respectively

t_1=(x/2)/3=x/6                                      …(i)

x_1=4.5 t_2 and x_2=7.5 t_2

So, x_1+x_2=x/2⇒4.5 t_2+7.5 t_2=x/2

t_2=x/24                                              …(ii)

Total time t=t_1+2t_2=x/6+x/12=x/4

So, average speed =4 m/sec

Q8. If the distance covered is zero, then displacement

•  Must be zero
•  May or may not be zero
•  Cannot be zero
•  Depends upon the particle

Solution

Magnitude of displacement can’t exceed distance

Q9.

Two trains, each travelling with a speed of 37.5 kmh^(-1), are approaching each other on the           same straight track. A bird that can fly at 60 km/h files off from one train when they are 90       km apart and heads directly for the other train. On reaching the other train, it files back to the         first and so on. Total distance covered by the bird is

•  90 km
  
•  54 km
•  36 km
•  72 km
  

Solution

Relative speed of trains =37.5+37.5=75 kmh^(-1)

Time taken by the trains to meet =90/75=6/5 h

Since speed of bird =60 kmh^(-1), distance travelled by the bird =60×6/5=72 km

Q10. For motion of an object along the x-axis, the velocity v depends on the displacement  x as v=3x^2-2x, then what is the acceleration at x=2 m

•  48 ms^(-2)
•  80 ms^(-2)
•  18 ms^(-2)
•  10 ms^(-2)

Solution

Given v=3x^2-2x; differentiating v, we get

dv/dt=(6x-2)  dx/dt=(6x-2)v

⇒a=(6x-2)(3x^2-2x) Now put x=2 m

⇒a=(6×2-2)(3(2)^2-2×2)=80 ms^(-2)