Physics Quiz-6
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.
An engine of a train moving with uniform acceleration passes an electric pole with velocity u and the last compartment with velocity v. The middle part of the train passes past the same pole with a velocity of
Solution
v^2-u^2=2as
Suppose velocity when middle part passes =v_m
Then v_m^2-u^2=2as×1/2=as
or v_m^2=u^2+as=u^2+(v^2-u^2)/2=(u^2+v^2)/2
⇒v_m=√((u^2+v^2)/2)
Q2.A body dropped from the top of a tower covers a distance 7x in the last second of its journey, where x is the distance covered in the first second. How much time does it take to reach the ground?
Solution
Given 7x=g/2(2n-1) and x=1/2 g(1)^2
Solving these two equations, we get n=4 s
Q3. The deceleration experienced by a moving motor boat, after its engine is cut-off is given by dv/dt=-kv^3, where k is constant. If v_0 is the magnitude of the velocity at cut-off, the magnitude of the velocity at a time t after the cut-off is
Solution
Here dv/dt=-kv^3
or dv/v^3 =-kdt or ∫_(v_0)^v▒dv/v^3 =∫_0^t▒〖-k dt〗
or [-1/(2v^2 )]_(v_0)^v=-kt or -1/(2v^2 )+1/(2v_0^2 )=-kt
or v^2=(v_0^2)/(1+2v_0^2 kt) or v=v_0/√(2v_0^2 kt+1)
Q4. An elevator in which a man is standing is moving upwards with a speed of 10 ms^(-1). If the man drops a coin from a height of 2.45 m from the floor of elevator, it reaches the floor of the elevator after time (g =9.8 ms^(-2))
Solution
Let the initial relative velocity, relative acceleration and relative displacement of the coin with respect to the floor of the lift be u_r,a_r, and s_r, then
s_r=u_r t+(1/2) a_r t^2
and u_r=u_c-u_l=10-10=0
a_r=a_c-a_l=(-9.8)-0=-9.8 ms^(-2)
s_r=s_c-s_l=-2.45 m
-2.45=0(t)+(1/2)(-9.8) t^2
or t^2=1/2 or t=1/√2 s
Q5.The ratio of the average velocity of a train during a journey to the maximum velocity between two stations is
Solution
Since maximum velocity is more than average velocity, ratio of the average velocity to maximum velocity has to be less than one.
Q6.
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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 |
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
Q7.
Between two stations a train accelerates from rest uniformly at first, then moves with constant,
and finally retards uniformly to come to rest. If the ratio of the time taken is 1: 8: 1 and the
maximum speed attained be 60 kmh^(-1), then what is the average speed over the whole journey?
Solution
S_1=1/2 at^2=1/2 (at)t=60t/2=30t
S_2=60×8 t=480 t,S_3=S_1=30t
v_av=(S_1+S_2+S_3)/(t+8t+t)=54 km h^(-1)
Q8.
A body is dropped from a height 39.2 m. After it crosses half distance, the acceleration due to gravity ceases to act. The body will hit the ground with velocity (Take g =10 ms^(-2)):
Solution
Suppose v be the velocity of the body after falling through half the distance. Then
s=39.2/2=19.6 m,u=0 and g=9.8 ms^(-2)
v^2=u^2+2gh=0^2+2×9.8×19.6 v=19.6 ms^(-1)
When the acceleration due to gravity ceases to act, the body travels with the uniform velocity of 19.6 ms^(-1). So it hits ground with velocity 19.6 ms^(-1)
Q9.
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The acceleration-time graph of a particle moving along a straight line is as shown in Fig. At what time the particle acquires its initial velocity?  |
Q10. The graph below describes the motion of a ball rebounding from a horizontal surface being released from a point above the surface. Assume the ball collides each time with the floor inelasticity. The quantity represented on the y-axis in the ball’s (take upward direction as positive
Solution
We know that for a body thrown up, its displacement is given as S=ut-1/2 gt^2. So the s-t graph is parabolic downwards.
Also the ball collides inelastically, so it will rebound to less height every time as shown in the graph
t_1,t_2,t_3 are the instants when the ball collides with ground. Here slope of the s-t graph is suddenly changing from negative to positive. It means velocity before collision is negative (downwards) and after collision is positive (upwards)
