Download App

PHYSICS [5 Mark Question Answer]

Laws of Motion · ICSE STD 9 (ICSE - English Medium). 11 questions.

Questions

[5 Mark Question Answer]

🎯

Test yourself on this topic

11 questions · timed · auto-graded

Question 15 Marks
A stone is dropped from a tower $98 $m high. With what speed should a second stone be thrown $1$ s later so that both hit the ground at the same time?
Answer
Height of tower $=98 m$
Acceleration due to gravity on stone $=9.8 ms^{-2}$.
Initial speed of ball $=0 ms^{-1}$.
Let initial speed of second stone is $v ms ^{-1}$.
We know from second equation of motion
$S = ut + 1/2 a Xt^2.$
$98 = 0 + 1/2 X9.8Xt^2.$
$t^2 = 20$
$t= 4.47 \sec.$
As second stone is thrown $1$ sec later so time taken by second body to cover distance of $98 m$ is $= 4.47 - 1 = 3.47sec.$
So again put t= 3.47 sec and S = 98 m in second equation of motion we get
$98 = vX3.47 + 1/2 X9.8X3.47X3.47.$
$98 = 3.47Xv + 59$
$3.47X v = 98 - 59$
$v= 39/3.47 = 11.23 ms^{-1}.$
Initial speed of second stone should be $11.23 ms^{-1}.$
View full question & answer→
Question 25 Marks
A stone is thrown vertically upward with a velocity of $9.8$ m/s. When will it reach the ground?
Answer
Initial speed of ball is $=9.8 ms^{-1}$.
Acceleration due to gravity $=-9.8 ms^{-2}$.
Final speed at maximum height $=0 ms^{-1}$.
We know $v=u+a t$
$0=9.8-9.8 t$
$T=1 sec .$
We know $v^2-u^2=2$ as
At highest point final velocity is zero so
$0-9.8 \times 9.8=2 \times(-9.8) S$
$S=4.9 m .$
for highest point initial velocity is zero
Acceleration due to gravity is $=9.8 ms^{-2}$.
Final velocity at ground is v
$\text { V2 - } 0=2 \times 9.8 \times 4.9$
$V=9.8 ms^{-1}$
Time taken to reach ground from highest point
$V=u+a t$
$9.8=0+9.8 t$
$T=9.8 / 9.8=1 sec$
Total time $=$ time of ascent + time of descent.
Total of flight = $1+1$ = 2 seconds.
View full question & answer→
Question 35 Marks
Below Figure shows the velocity-time graph of a particle of mass 200 g moving in a stra ight line. Calculate the force acting on the particle.
Answer

As we know that slope of velocity time graph gives acceleration so acceleration of the body is$= (v - u)/ ( t_2 - t_1) ={20 - 0)/(5-0) = 4 ms^{-2}$​​​​​​​
Mass of the body is$= 200 g =0.2 kg$
Force applied is equal to $= 0.2 x4 = 0.8 N.$
View full question & answer→
Question 45 Marks
What do you mean by the conservation of momentum? Briefly, explain the $c_o$llision between two bodies and the conservation of momentum.
Answer
Conservation of momentum in case of a collision between two bodies means the total momentum before and after collision remains unchanged or conserved, provided no net force acts on the system.

Consider two bodies A and 8 having masses m 1 and m , and initial velocities u 1 and u , respectively. The bodies collide head on with each other and their collision lasts for $t$ seconds. Suppose the velocities of $A$ and 8 after collision are $v 1$ and $v$, respectively. Assume that no external forces are acting on the bodies.
Rate of change of momentum of ball $A=m_1\left(v_1-u_1\right) / t$.
Rate of change of momentum of ball $8=m_1\left(v_1-u_1\right) / t$.
If $F_{A B}$ is the force exerted by $A$ on 8 and $F_{B A}$ is the force ex
Rated by 8 on A , we can write
Rated by 8 on A, we can write
$F_{AB}= m_1(v_1 - u_1)/t .$
$F_{BA} = m_1 (v_1 - u_1)/t .$
$F_{AB}= -F_{BA}$
$m_1(v_1 - u_1)/t = - m_2(v_2- u_2)/t .$
$m_1(v_1 - u_1) = - m_2(v_2- u_2)$
$m_1v_1 +m_2v_2 = m_1u_1+m_2u_2$
So total momentum after collision = total momentum before collision.
This proves conservation of momentum during collision.
View full question & answer→
Question 55 Marks
Show that the rate of change of momentum = mass × acceleration. Under what condition does this relation hold?
Answer
Let a force 'F' be applied on a body of mass m for a time 't' due to which its velocity changes from u to v. Then,
Initial momentum of body = mu
Final momentum of body = mv
Change in momentum of the body in 't' seconds = mv -­­ mu = m (v - u)
Rate of change of momentum = Change in momentum/time
= [m (v - u)]/t
However, acceleration a = Change in velocity/time = (v - u)/t
Therefore, rate of change of momentum = ma = mass × acceleration
This relation holds true when the mass of the body remains constant.
View full question & answer→
Question 65 Marks
Two bodies $P$ and $Q$, of same masses m and 2m are moving with velocities 2 v and v respectively. Compare their
(i) Inertia, (ii) Momentum, and (iii) The force required to stop them in the same time.
Answer
Mass of $P$ is $m_1=m$.
Velocity of P is $v _1=2 v$
Mass of $Q$ is $m_2=2 m$
Velocity of $Q$ is $v_2=v$.
(i) inertia of P/inertia of $Q=m_1 / m_2=1 / 2$.
So ratio of inertia of two bodies is $1: 2$.
(ii) Momentum of $P /$ momentum of $Q = m _1 v _1 / m _2 v _2=1$
So ratio of momentum of two bodies is $1: 1$.
(iii) As force required to stop them is equal to change in their momentum from moving to rest.
So ratio would be same as the ratio of their momentum i.e $1: 1$.
View full question & answer→
Question 75 Marks
The distance-time values for an object moving along straight line are given below:
Time (s) Distance (m)
$0$ $0$
$1$ $1$
$2$ $8$
$3$ $27$
Answer
Time (s) Distance (m) Velocity $(ms^{-1})=$ distance/time Acceleration $(ms^{-1})= (v-u)/(t_2-t_{1)}$
$0$ $0$ $0$ $0$
$1$ $1$ $1$ $1$
$2$ $8$ $4$ $3$
$3$ $27$ $9$ $5$
So the object moves with increasing acceleration
View full question & answer→
Question 85 Marks
The speed-time graph of a body having a mass of $100 g$ moving along a straight line path is shown here. Find the force acting on the body.
Answer

As we know slope or speed time graph gives acceleration so we can find the acceleration of the body.
$a=(v-u)/t= (10-0) /( 4-0)= 5/2 = 2.5ms^{-2}$​​​​​​​
Massof the body is $= 100 g= 0.1kg$
Forece= mass x acceleration.
Force$= 0.1 x 2.5 = 0.25 N$
View full question & answer→
Question 95 Marks
Show that gravity decreases at higher altitudes.
Answer
We know that the law of gravitation is
$F = G ( m_1 x m_2)/R^2$
Here the F is the force due to attraction and this force is equal to the weight of the body $m_2g.$
So$ m_2g = G ( m_1 x m_2)/R^2$
$g= (G x m_2)/R^2.$
Here R is the distance between the earth centre and the object centre. Now if we go on higher altitude say ‘H’ then this R would increase to (R + H)
And the value of gravity at height H becomes
$g’= (G x m_2)/( R + H)^2.$
As denominator increases so g’ would be less than g and hence we can say that gravity decreases at higher altitudes.
View full question & answer→
Question 105 Marks
At what height above the earth's surface would the value of acceleration due to gravity be half of what it is on the surface? Take the radius of earth to be R.
Answer
As we know
$g=G \times \frac{M}{R^2}$
So let at height $H$ the value of $g$ is half that of the earth earth surface then $g$ at $R + H$ would be equal to
$g^{\prime}=G \times \frac{M}{(R+H)^2}$
Now $g ^{\prime} / g =1 / 2$
$1 / 2=R^2 /(R+H)^2$
$(R+H)^2=2 R^2$
$R+H=\sqrt{ } 2 R$
$H=(\sqrt{ } 2-1) R$
View full question & answer→
Question 115 Marks
Write the answer of the question with reference to laws of gravitation.State the universal law of gravitation.
Answer
The universal law of gravitation states that every object in the universe attracts every other object with a force called the gravitational force. The force acting between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
For two objects of masses, $m_1$ and $m_2$ and the distance between them $r$, the force ( $F$ ) of attraction acting between them is given by the universal law of gravitation as:-
$F = G \frac{ m _1 m _2}{ r ^2}$
Where $G$ is the universal gravitation constant given by:
$G =6.67 \times 10^{-11} Nm ^2 kg ^{-2}$
View full question & answer→
Generate Paper FreeAll Laws of Motion topics
[5 Mark Question Answer] - PHYSICS STD 9 Questions - Vidyadip