Quiz Solutions
1. Mass, m = 4 kg
Initial velocity, u = 0
Force, F = 16 N
Therefore acceleration, a = F/m = 16/4 = 4 m/s2
Time, t = 10 sec
Therefore final velocity, v = u + at = 0 + 4*10 = 40 m/s
So, final kinetic energy = ½ mv2 = 4*40*40/2 = 3200 J
Option (b)
2. Mass of each block, m =5 kg
Speed of each block, u = 2 m/s
1
1
Therefore, initial kinetic energy of system is, 𝐾𝑖 = 2 (2 𝑚𝑢2 ) = 2 (2 ∗ 5 ∗ 22 ) = 20 𝐽
Finally both blocks come to rest.
Therefore, final kinetic energy of system is, Kf = 0
Hence, Work done is, W = Kf – Ki = 0 – 20 = -20 J
Since, there is no external force acting on the system so work done by external forces is zero
and total work done is done by internal forces.
Hence, Wint = -20J
Option (a)
3. Initial momentum of body = Pi
Momentum increases by 20%
Therefore, final momentum, Pf = 20% of Pi + Pi
20
6
= 100 𝑃𝑖 + 𝑃𝑖 = 5 𝑃𝑖
𝑃2
If the mass of the body is ‘m’, then Kinetic energy of body is, 𝐾 = 2𝑚
𝑃𝑓2
36
𝑃2
𝑖
Now final kinetic energy is, 𝐾𝑓 = 2𝑚 = 25 ∗ 2𝑚
𝑃2
𝑖
and initial kinetic energy is, 𝐾𝑖 = 2𝑚
11
𝑃2
𝑖
Change in kinetic energy, ∆𝐾 = 𝐾𝑓 − 𝐾𝑖 = 25 ∗ 2𝑚
Percentage increase in Kinetic energy is =
∆𝐾
𝐾𝑖
∗ 100 =
2
11 𝑃𝑖
∗
25 2𝑚
𝑃2
𝑖
∗ 100 =
2𝑚
Option (c)
4. Initial kinetic energy of block = Ki
Kinetic energy increases by 300%
Therefore, final kinetic energy is Kf = 3Ki + Ki = 4Ki
If the mass of the body is ‘m’, then momentum of the body is, P = √2mK
Final momentum of body is, Pf = √2𝑚𝐾𝑓 = √2𝑚 ∗ 4𝐾𝑖 = 2√2𝑚𝐾𝑖
11
25
∗ 100 = 44%
Initial momentum of body is, 𝑃𝑖 = √2𝑚𝐾𝑖
Change in momentum is, ∆𝑃 = 𝑃𝑓 − 𝑃𝑖 = 2√2𝑚𝐾𝑖 − √2𝑚𝐾𝑖 = √2𝑚𝐾𝑖
Percentage change in momentum is =
∆𝑃
𝑃𝑖
∗ 100 =
√2𝑚𝐾𝑖
√2𝑚𝐾𝑖
∗ 100 = 100%
Option (c)
5. Ball falls with a velocity ‘u’ from height 10 m.
1
Total mechanical energy at height 10m = 2 𝑚𝑢2 + 10𝑚𝑔
where ‘m’ and ‘g’ are mass of ball and acceleration due to gravity respectively.
At ground it loses 50% of its total energy.
1 1
Then total mechanical energy at ground is = 2 (2 𝑚𝑢2 + 10𝑚𝑔)
Ball rebounces to a height of 10 m.
Total energy at height 10 m = 10gh
Total energy at height 10 m should be equal to total energy at ground.
1 1
∴ 10𝑚𝑔 = ( 𝑚𝑢2 + 10𝑚𝑔)
2 2
𝑢2
10 ∗ 2𝑔 =
+ 10𝑔
2
10𝑔 =
𝑢2
2
∴ 𝑢 = √20𝑔 = √20 ∗ 9.8 = 14𝑚/𝑠
Option (d)
6.
The bullet gets embedded into the bob and then both of them move together.
This tells that the collision is an inelastic one so, only linear momentum is conserved here
and kinetic energy is not conserved.
Before the collision, only bullet is in motion. Let its mass and velocity be ‘m’ and ‘u’
respectively.
Then, initial momentum is = mu
where m = 0.01 kg
u = 2*10-2 m/s
After the collision, bullet gets embedded into the bob and both of them move together.
Let the mass of bob be ‘M’ and their (bob + bullet) combined velocity just after the collision
be ‘v’.
Then, final momentum is = (M+m)v
where M = 1 kg
Now, initial momentum = final momentum
mu = (M+m)v
∴v=
𝐦𝐮
𝐌+𝐦
=
𝟎.𝟎𝟏∗𝟐∗𝟏𝟎−𝟐
𝟏+𝟎.𝟎𝟏
= 1.98
m
s
Let the bob rises to a vertical height ‘h’ before swinging back.
Then potential energy at highest point is equal to the kinetic energy just after collision
1
(M + m)gh = (M + m)v 2
2
v2
1.982
h=
=
= 0.2m
2g 2 ∗ 9.8
Option (b)
7.
Total downward force is = mg + f
where m = mass of load = 1800 kg
f = frictional force = 4000 N
Elevator should apply a force equal to total downward force in upward direction.
Speed of elevator, v = 2 m/s
Therefore power supplied by the motor = (mg + f)v
= (1800*10 + 4000)*2
= 44000 W
Option (b)
8. Initially ball is at height 3.6 m and dropped from there.
So, velocity at which ball hits the horizontal surface is = √2ghi
= √2 ∗ 9.8 ∗ 3.6 = 8.4 m/s
Ball rebounds to a height of 1.6 m.
So, velocity at which ball leaves surface = √2ghf
= √2 ∗ 9.8 ∗ 1.6 = 5.6 m/s
Coefficient of restitution =
𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐨𝐟 𝐬𝐞𝐩𝐞𝐫𝐚𝐭𝐢𝐨𝐧
𝐕𝐞𝐥𝐨𝐜𝐢𝐭𝐲 𝐨𝐟 𝐚𝐩𝐩𝐫𝐨𝐚𝐜𝐡
=
𝟓.𝟔
𝟖.𝟒
=
𝟐
𝟑
Option(a)
9. For first particle
mass, m1 = 0.001 kg
velocity, v1 = 3i – 2j
For second particle
mass, m2 = 0.002 kg
velocity, v2 = 4j – 6k
Collision is a perfectly inelastic one.
Let velocity of the combined particle be v
Using conservation of momentum
Momentum before collision = Momentum after collision
m1v1 + m2v2 = (m1 + m2)v
0.001(3i – 2j) + 0.002(4j – 6k)=(0.001 + 0.002)v
v = i + 2j – 4k
Option (a)
10.
Let mass of each ball be ‘m’.
Their initial velocities be u1 and u2 and their final velocities be v1 and v2 respectively.
Here u1 = 9m/s
u2 = 0
Along x axis
Total momentum before collision = mu1 = 9m
Total momentum after collision = mv1cos30⁰ + mv2cos30⁰
√𝟑
𝟐
= (v1 + v2)m*
Applying conservation of momentum we get
9m = (v1 + v2)m*
√𝟑
𝟐
v1 + v2 = 6√3
........(i)
Along y axis
Total momentum before collision = 0
Total momentum after collision = mv1sin30⁰ - mv2sin30⁰
𝟏
= (v1 - v2)m*
𝟐
Applying conservation of momentum we get
𝟏
0 = (v1 - v2)m*
𝟐
v1 - v2 = 0
By solving eq. (i) and (ii) we get
v1 = v2 = 3√3
........(ii)
m
s
𝟏
𝟏
𝟐
𝟐
Total kinetic energy before collision = 𝐦𝐮𝟐𝟏 =
𝟏
𝟏
𝟐
𝟏
𝟐
∗ 𝐦 ∗ 𝟗𝟐 = 𝟒𝟎. 𝟓𝐦
Total kinetic energy after collision = 𝐦𝐮𝟐𝟏 + 𝐦𝐮𝟐𝟐
𝟐
𝟏
= ∗ 𝐦 ∗ (𝟑√𝟑) + ∗ 𝐦 ∗ (𝟑√𝟑)
𝟐
𝟐
𝟐
= 𝟐𝟕𝐦
Clearly total kinetic energy before the collision is not equal to the total kinetic energy after
the collision.
Hence, conservation of kinetic energy is not valid.
Option (d)