StudentAidUnit 2: Force and Motion — Numericals
11th Class Physics · Unit 2: Force and Motion
2.1.The magnitude of cross and scalar products of two vectors are 4\sqrt{3} and 4, respectively. Find the angle between the vectors.
Given
Cross product magnitude
|vec{A} × vec{B}| = 4sqrt{3}
Scalar product magnitude
|vec{A} · vec{B}| = 4
Formula
Cross product formula
|vec{A} × vec{B}| = AB sinθ
Equation (1)
4sqrt{3} = AB sinθ
Scalar product formula
|vec{A} · vec{B}| = AB cosθ
Equation (2)
4 = AB cosθ
Dividing Eq. (1) by Eq. (2)
frac{4sqrt{3}{4} = ABsinθABcosθ
Simplification
sqrt{3} = tanθ
Result
Angle
θ = tan-1(sqrt{3}) = 60°
2.2.A helicopter is ascending vertically at the rate of 19.6 ms^{-1}. When it is at a height of 156.8 m above the ground, a stone is dropped. How long does the stone take to reach the ground?
Given
Initial velocity of stone (upward)
vi = 19.6 m/s
Initial height of stone
h = 156.8 m
Acceleration
a = -g = -9.8 m/s2
Vertical motion (ascending phase)
vf = vi + at
At highest point
0 = 19.6 - 9.8 · t1
Result
Time to reach maximum height
t1 = 2 sec
Height gained during ascent
h = vit + 12at2
Substitution
h = 19.6 × 2 - 12 × 9.8 × 4
Calculation
h = 39.2 - 19.6 = 19.6 m
Net height above ground
Net height = 156.8 + 19.6 = 176.4 m
Descent phase from max height
vi = 0, h = 176.4 m, a = 9.8 m/s2
Using 2nd equation of motion
h = vit2 + 12gt22
Substitution
176.4 = 0 + 12 × 9.8 × t22
Calculation
176.4 = 4.9 · t22
Solving for t₂
t22 = 176.44.9 = 36
Descent time
t2 = 6 sec
Total time
t = t1 + t2 = 2 + 6 = 8 sec
2.3.If |\vec{A} + \vec{B}| = |\vec{A} - \vec{B}|, then prove that \vec{A} and \vec{B} are perpendicular to each other.
Given
Condition
|vec{A} + vec{B}| = |vec{A} - vec{B}|
To prove
vec{A} perp vec{B} (i.e., θ = 90°)
Taking square on both sides
|vec{A} + vec{B}|2 = |vec{A} - vec{B}|2
Expansion
A2 + B2 + 2vec{A} · vec{B} = A2 + B2 - 2vec{A} · vec{B}
Simplification
4vec{A} · vec{B} = 0
Result
Result
vec{A} · vec{B} = 0
Using dot product formula
AB cosθ = 0
Angle between vectors
cosθ = 0 ⇒ θ = 90°
Conclusion
Hence, vec{A} is perpendicular to vec{B}
2.4.A body of mass M at rest explodes into 3 pieces, two of which of mass M/4 each are thrown off in perpendicular directions with velocities of 3 ms^{-1} and 4 ms^{-1} respectively. Find the velocity of the 3rd piece with which it will be flown away.
Given
Mass of 1st piece
m1 = M4
Mass of 2nd piece
m2 = M4
Mass of 3rd piece
m3 = M - (M4 + M4) = M2
Velocity of 1st piece (x-axis)
v1 = 3 m/s
Velocity of 2nd piece (y-axis)
v2 = 4 m/s
Momentum of 1st piece
vec{p}1 = M4 × 3 = 3M4 (along x-axis)
Momentum of 2nd piece
vec{p}2 = M4 × 4 = M (along y-axis)
Total momentum of two pieces
vec{P}_{total = sqrt{(3M4)2 + M2
Calculation
vec{P}_{total = sqrt{frac{9M2{16} + M2 = sqrt{frac{9M2 + 16M2{16} = sqrt{frac{25M2{16} = 5M4
By conservation of momentum
vec{P}3 = -vec{P}_{total
Momentum of 3rd piece
m3 × v3 = 5M4
Substitution
M2 × v3 = 5M4
Result
Velocity of 3rd piece
v3 = 5M4 × 2M = 52 = -2.5 m/s
2.5.A cricket ball is hit upward at an angle of 45° with velocity of 20 ms^{-1}. Find its: (a) time of flight (b) maximum height (c) how far away it hits the ground
Given
Angle of projection
θ = 45°
Initial velocity
vi = 20 m/s
Part (a): Time of flight
t = 2visinθg = 2 × 20 × sin 45°9.8
Calculation
t = 40 × 0.7079.8 = 28.289.8 = 2.9 s
Part (b): Maximum height
H = vi2sin2θ2g = frac{(20)2(sin 45°)2{2 × 9.8}
Calculation
H = frac{400 × (0.707)2{19.6} = 400 × 0.519.6 = 20019.6 = 10.20 m
Part (c): Horizontal range
R = vi2sin 2θg = (20)2 sin 2(45°)9.8
Calculation
R = 400 × sin 90°9.8 = 400 × 19.8 = 40.8 m ≈ 41 m
(a) t = 2.9 s; (b) H = 10.20 m; (c) R = 41 m
2.6.A 20g ball hits the wall of a squash court with a constant force of 50 N. If the time of impact of force is 0.50 s, find the impulse.
Given
Mass of ball
m = 20g = 201000 kg = 0.020 kg
Force applied
F = 50 N
Time of impact
Δt = 0.50 s
Formula
Impulse formula
Impulse = F × Δt
Working
Calculation
Impulse = 50 × 0.50 = 25 N s
Impulse = 25 N s
2.7.A ball is kicked by a footballer. The average force on the ball is 240 N, and the impact lasts for a time interval of 0.25 s. Calculate: (a) change in momentum (b) state direction of change in momentum.
Given
Force applied
F = 240 N
Time interval
Δt = 0.25 s
Part (a): Change in momentum
Δp = F × Δt = 240 × 0.25 = 60 N s
Part (b): Direction of change in momentum
The direction of change in momentum is the same as the applied force.
(a) Δp = 60 N s; (b) Same as applied force direction
2.8.An aeroplane moving horizontally at a speed of 200 m s^{-1} at a height of 8 km to drop a bomb on a target. Find horizontal distance from the target should the bomb be released.
Given
Initial horizontal velocity
vi = 200 m/s
Height
y = 8 km = 8000 m
Initial vertical velocity
viy = 0
Acceleration
a = g = 9.8 m/s2
Time to reach ground using 2nd equation of motion
S = vit + 12at2
Substitution
8000 = 0 + 12 × 9.8 × t2
Calculation
8000 = 4.9t2
Solving for t
t2 = 80004.9 = 1632.65
Time calculation
t = sqrt{1632.65} ≈ 40.4 s
Horizontal distance
x = vi × t = 200 × 40.4 = 8080 m = 80801000 km = 8.08 km
x = 8080 m = 8.08 km
2.9.Why does range of a projectile remain the same when angle of projection is changed from \theta to \theta' = 90 - \theta. Also show that for complementary angles of projection the ratio R/R' is equal to 1.
Given
First angle
1st angle = θ
Second angle
2nd angle = θ' = 90° - θ
Part (a): Show that ranges are equal
R = vi2 sin 2θg quad ...... (1)
For complementary angle
R' = vi2 sin 2(90° - θ)g
Expansion
R' = vi2 sin(180° - 2θ)g
Using sin identity
R' = vi2 sin 2θg quad ...... (2)
Comparing Eq. (1) and (2)
R = R'
Part (b): Show that for complementary angles
θand (90° - θ) are complementary angles. So, dividing Eq. (1) by (2)
Ratio calculation
RR' = frac{vi2 sin 2θg{vi2 sin 2θg = 1
R = R' and RR' = 1
2.10.A trolley of mass 1.0 kg moving with velocity 1.0 m s^{-1} collides with a similar trolley at rest: (i) After collision, the 1st trolley comes to rest whereas the second start moving with velocity of 1.0 m s^{-1} in the same direction. Show that it is an example of an elastic collision. (ii) After the collision, they stick together and move away with a velocity of 0.5 ms^{-1}. Show that it is an example of an inelastic collision.
Given
Mass of both trolleys
m1 = m2 = 1 kg
Initial velocity of 1st trolley
v1 = 1 m/s
Initial velocity of 2nd trolley
v2 = 0
Part (i): Elastic collision case
v1' = 0, quad v2' = 1 m/s
K.E before collision
K.E = 12m1v12 + 12m2v22
Substitution
K.E = 12(1)(1)2 + 12(1)(0)2 = 0.5 J
K.E after collision
K.E' = 12m1v1'2 + 12m2v2'2
Substitution
K.E' = 12(1)(0)2 + 12(1)(1)2 = 0.5 J
Observation
K.E before collision = K.E after collision
Law of conservation of K.E. holds
Therefore, it is an example of elastic collision.
Part (ii): Inelastic collision case
After collision, they stick together and move with velocity v' = 0.5 m/s
K.E before collision
K.E = 12(1)(1)2 + 12(1)(0)2 = 0.5 J
K.E after collision
K.E' = 12(m1 + m2)v'2
Substitution
K.E' = 12(2)(.5)2 = 0.25 J
Observation
K.E before collision neq K.E after collision
Conclusion
Therefore, it is an example of inelastic collision.
(i) Elastic collision - K.E conserved; (ii) Inelastic collision - K.E not conserved
2.11.A railway wagon of mass 4 × 10^{4} kg moving with velocity of 3 ms^{-1} collide with another wagon of mass 2 × 10^{4} kg velocity is at rest. They stick together and move off together. Find their combined velocity.
Given
Mass of first wagon
m1 = 4 × 104 kg
Velocity of first wagon
v1 = 3 m/s
Mass of second wagon
m2 = 2 × 104 kg
Velocity of second wagon
v2 = 0 m/s
Applying law of conservation of momentum
m1v1 + m2v2 = (m1 + m2)vf
Substitution
(4 × 104)(3) + (2 × 104)(0) = (6 × 104)vf
Calculation
12 × 104 = 6 × 104 vf
Result
Final velocity
vf = 2 m s-1
2.12.A car with mass 575 kg moving at 15.0 ms^{-1} smashes into the rear end of a car with mass 1575 kg moving at 5 ms^{-1} in the same direction. (i) What is the final velocity if the wrecked car locks together? (ii) How much kinetic energy is lost in the collision?
Given
Mass of first car
m1 = 575 kg
Velocity of first car
v1 = 15 m/s
Mass of second car
m2 = 1575 kg
Velocity of second car
v2 = 5 m/s
Part (i): Final velocity
m1v1 + m2v2 = (m1 + m2)v'
Substitution
575(15) + 1575(5) = 2150v'
Calculation
8625 + 7875 = 2150v'
Sum
16500 = 2150v'
Result
Final velocity
v' = 165002150 = 7.67 m/s
Part (ii): Loss in kinetic energy
Loss in K.E = K.Ei - K.Ef ...... (1)
Initial K.E.
K.Ei = 12m1v12 + 12m2v22
Substitution
K.Ei = (575)(15)2 + (1575)(5)2
Calculation
K.Ei = 64687.5 + 19687.5 = 84375 J
Final K.E.
K.Ef = 12(m1 + m2)v'2
Substitution
K.Ef = 12(575 + 1575)(7.67)2
Calculation
K.Ef = 12(2150)(58.83) = 63238.75 J
Loss in K.E
Loss = 84375 - 63238.75 = 21136.25 J