Class 11 Physics Assignment

Solution: Given resultant velocity v = 80 km/h and one rectangular component vx = 40 km/h. For perpendicular components, v² = vx² + vy². Therefore vy = √(v² − vx²) = √(80² − 40²) = √(6400 − 1600) = √4800 = 40√3 km/h = 69.28 km/h. Hence, the other component is 69.28 km/h.

Solution: For the first ball, vx = 1 m/s and vy = √3 m/s. Therefore tanθ₁ = vy/vx = √3/1 = √3, so θ₁ = 60°. For the second ball, vx = 2 m/s and vy = 2 m/s. Therefore tanθ₂ = 2/2 = 1, so θ₂ = 45°. The angle between their paths is θ₁ − θ₂ = 60° − 45° = 15°.

Solution: The component of a vector A along x-axis is Ax = A cosθ and along y-axis is Ay = A sinθ. Here A = 10 units and θ = 30°. So Ax = 10 cos30° = 10 × √3/2 = 5√3 units. Also Ay = 10 sin30° = 10 × 1/2 = 5 units.

Solution: If the rectangular components are Ax = 6 units and Ay = 8 units, then magnitude A = √(Ax² + Ay²) = √(6² + 8²) = √100 = 10 units. Direction with x-axis is given by tanθ = Ay/Ax = 8/6 = 4/3. Hence θ = tan⁻¹(4/3) ≈ 53.1°.

Solution: For two forces F1 and F2 making angle θ, resultant R = √(F1² + F2² + 2F1F2 cosθ). Here F1 = 5 N, F2 = 10 N, θ = 30°. R = √(25 + 100 + 2×5×10×cos30°) = √(125 + 100×√3/2) = √(125 + 50√3). Numerically, R ≈ √211.6 = 14.56 N.

Solution: Let each vector have magnitude A and angle between them be θ. Resultant R is also A. Using R² = A² + A² + 2A²cosθ, we get A² = 2A² + 2A²cosθ. Dividing by A², 1 = 2 + 2cosθ. Hence cosθ = −1/2 and θ = 120°.

Solution: Using resultant formula R = √(F1² + F2² + 2F1F2 cosθ). Here F1 = 5 N, F2 = 10 N and θ = 120°. R = √(25 + 100 + 2×5×10×cos120°). Since cos120° = −1/2, R = √(125 − 50) = √75 = 5√3 = 8.66 N.

Solution: The general expression for resultant of two vectors is R² = A² + B² + 2ABcosθ. Given R² = A² + B². Comparing both equations, 2ABcosθ = 0. Since A and B are non-zero, cosθ = 0. Therefore θ = 90°.

Solution: For perpendicular velocity components vx and vy, resultant speed is v = √(vx² + vy²). Here vx = 3 m/s and vy = 4 m/s. Therefore v = √(3² + 4²) = √(9 + 16) = √25 = 5 m/s.

Solution: East and north displacements are perpendicular. Resultant displacement S = √(12² + 5²) = √(144 + 25) = √169 = 13 m. Direction is tanθ = 5/12, so θ = tan⁻¹(5/12) north of east.

Solution: For horizontal projection, vertical initial velocity is zero. Vertical displacement h = 1/2 gt². Given h = 80 m and g = 10 m/s². So 80 = 1/2 × 10 × t² = 5t². Thus t² = 16 and t = 4 s.

Solution: In horizontal projection, horizontal velocity remains constant. From Q11, time of flight t = 4 s. Horizontal range R = horizontal velocity × time = 20 × 4 = 80 m.

Solution: Time of flight for projectile is T = 2u sinθ/g. Here u = 40 m/s, θ = 30° and g = 10 m/s². T = 2×40×sin30°/10 = 80×1/2/10 = 40/10 = 4 s.

Solution: Maximum height H = u²sin²θ/2g. Here u = 40 m/s, θ = 30°, g = 10 m/s². H = 40² × (sin30°)² / 20 = 1600 × (1/2)² / 20 = 1600 × 1/4 / 20 = 400/20 = 20 m.

Solution: Horizontal range R = u²sin2θ/g. Here u = 40 m/s and θ = 30°. So R = 40² sin60° / 10 = 1600 × √3/2 / 10 = 80√3 m ≈ 138.6 m. The earlier value 160√3 was incorrect; correct range is 80√3 m.

Solution: Range of projectile on horizontal ground is R = u²sin2θ/g. For fixed u and g, R is maximum when sin2θ is maximum. The maximum value of sine is 1, so 2θ = 90°. Therefore θ = 45°.

Solution: Range is R = u²sin2θ/g. For θ = 30°, sin2θ = sin60°. For θ = 60°, sin2θ = sin120°. Since sin60° = sin120°, both ranges are equal.

Solution: Time of flight T = 2u sinθ/g. For the same speed, T is proportional to sinθ. Therefore T30:T60 = sin30°:sin60° = 1/2:√3/2 = 1:√3.

Solution: For projectile motion, maximum height H = u²sin²θ/2g and range R = u²sin2θ/g. Dividing, R/H = [u²sin2θ/g] / [u²sin²θ/2g] = 2sin2θ/sin²θ = 4cotθ. Given R/H = 80/20 = 4. Therefore 4cotθ = 4, so cotθ = 1 and θ = 45°.

Solution: Horizontal component is ux = u cosθ. Given u = 25 m/s and cos53° = 0.6. Therefore ux = 25 × 0.6 = 15 m/s.

Solution: Vertical component is uy = u sinθ. Given u = 25 m/s and sin53° = 0.8. Therefore uy = 25 × 0.8 = 20 m/s.

Solution: Initial speed is the resultant of horizontal and vertical components. u = √(ux² + uy²) = √(15² + 20²) = √(225 + 400) = √625 = 25 m/s.

Solution: Angle of projection is given by tanθ = uy/ux. Here uy = 20 m/s and ux = 15 m/s. Therefore tanθ = 20/15 = 4/3. Hence θ = tan⁻¹(4/3) ≈ 53.1°.

Solution: Centripetal acceleration is a = v²/r. Here v = 10 m/s and r = 2 m. Therefore a = 10²/2 = 100/2 = 50 m/s².

Solution: Relation between linear speed and angular speed is v = rω. Given r = 5 m and ω = 4 rad/s. Therefore v = 5 × 4 = 20 m/s.

Solution: Centripetal acceleration in terms of angular speed is a = rω². Given r = 5 m and ω = 4 rad/s. So a = 5 × 4² = 5 × 16 = 80 m/s².

Solution: For uniform angular motion, angular displacement θ = ωt. Given ω = 10 rad/s and t = 5 s. Therefore θ = 10 × 5 = 50 rad.

Solution: Frequency is number of revolutions per second. f = 20/10 = 2 Hz. Angular speed ω = 2πf = 2π × 2 = 4π rad/s.

Solution: Speed in uniform circular motion is circumference divided by time. v = 2πr/T. Given r = 7 m and T = 22 s. v = 14π/22. Taking π = 22/7, v = 14×22/(7×22) = 2 m/s.

Solution: Magnitude of vector A = 3i + 4j is |A| = √(3² + 4²) = √25 = 5.

Solution: Unit vector along A is A/|A|. For A = 3i + 4j, |A| = 5. Hence unit vector = (3i + 4j)/5 = (3/5)i + (4/5)j.

Solution: Dot product A·B = AxBx + AyBy. Here A = 2i + 3j and B = 4i − j. So A·B = 2×4 + 3×(−1) = 8 − 3 = 5.

Solution: Magnitude of cross product is |A×B| = AB sinθ. Given A = 6, B = 8 and θ = 30°. |A×B| = 6×8×sin30° = 48×1/2 = 24.

Solution: For two perpendicular vectors, resultant R = √(A² + B²). Here A = 3 and B = 4. Therefore R = √(3² + 4²) = √25 = 5.

Solution: When two vectors act in the same direction, resultant is the sum of magnitudes. Therefore R = 3 + 4 = 7.

Solution: When two vectors act in opposite directions, resultant magnitude is the difference of magnitudes. Therefore R = |4 − 3| = 1.

Solution: Let each component be a. Since components are perpendicular, resultant magnitude 20 satisfies 20² = a² + a² = 2a². Thus a² = 200 and a = √200 = 10√2 units.

Solution: Horizontal component of force is Fx = F cosθ. Given F = 10 N and θ = 60°. Fx = 10 cos60° = 10 × 1/2 = 5 N.

Solution: Vertical component of force is Fy = F sinθ. Given F = 10 N and θ = 60°. Fy = 10 sin60° = 10 × √3/2 = 5√3 N.

Solution: East and north velocities are perpendicular. Resultant speed v = √(5² + 12²) = √(25 + 144) = √169 = 13 m/s.

Solution: Rain has vertical velocity 10 m/s downward. Man has horizontal velocity 10 m/s. Apparent velocity of rain relative to man has horizontal component 10 m/s and vertical component 10 m/s. Therefore tanθ = 10/10 = 1, so θ = 45° with the vertical.

Solution: The swimmer’s velocity perpendicular to river flow is 5 m/s and river velocity is 3 m/s. These are perpendicular components. Resultant speed = √(5² + 3²) = √34 m/s.

Solution: In projectile motion, horizontal acceleration is zero. Therefore horizontal velocity remains constant throughout the motion. Hence vx = u cosθ.

Solution: Vertical velocity changes due to downward acceleration g. Taking upward as positive, vertical velocity after time t is vy = u sinθ − gt.

Solution: At the highest point, the projectile stops moving upward momentarily. Hence vertical component of velocity becomes zero.

Solution: At the highest point, vertical velocity is zero but horizontal velocity remains unchanged. Therefore speed at the highest point is u cosθ.

Solution: Projectile motion is symmetric for upward and downward journey on the same level. Time to reach maximum height is half the total time of flight. Therefore t = 6/2 = 3 s.

Solution: Horizontal velocity is constant and range = horizontal velocity × time of flight. Thus ux = range/time = 100/5 = 20 m/s.

Solution: For horizontal projection, horizontal velocity remains constant. Therefore horizontal velocity = horizontal distance/time = 60/3 = 20 m/s.

Solution: Velocity vector is v = 6i + 8j m/s. Speed = √(6² + 8²) = √100 = 10 m/s. Direction with x-axis is tanθ = 8/6 = 4/3, so θ ≈ 53.1°.

Solution:
If two rectangular components are perpendicular, then resultant velocity satisfies:
v² = v_x² + v_y²
Given v = 80 km h⁻¹ and v_x = 40 km h⁻¹.
So, v_y = √(80² − 40²) = √(6400 − 1600) = √4800 = 40√3.
Therefore, the other component is 40√3 km h⁻¹ or 69.3 km h⁻¹.

Solution:
Direction of velocity is given by tan θ = v_y/v_x.
For first ball: tan θ₁ = √3/1 = √3, so θ₁ = 60°.
For second ball: tan θ₂ = 2/2 = 1, so θ₂ = 45°.
Angle between their paths = 60° − 45° = 15°.

Solution:
Magnitude of A is:
|A| = √(3² + 4²) = √25 = 5.
Direction θ is given by:
tan θ = 4/3.
Therefore, θ = tan⁻¹(4/3) = 53.1° with the positive x-axis.

Solution:
Rectangular components are:
A_x = A cos θ = 25 × 4/5 = 20
A_y = A sin θ = 25 × 3/5 = 15
Hence the vector can be written as A = 20i + 15j.

Solution:
For two perpendicular vectors, resultant is:
R = √(A² + B²)
R = √(6² + 8²) = √(36 + 64) = √100 = 10 units.

Solution:
For two vectors:
R² = A² + B² + 2AB cos θ
Here A = B = R = 10.
100 = 100 + 100 + 200 cos θ
200 cos θ = −100
cos θ = −1/2
Therefore, θ = 120°.

Solution:
Dot product is:
A · B = A_xB_x + A_yB_y
A · B = 2 × 4 + 3 × (−1) = 8 − 3 = 5.

Solution:
First find dot product:
A · B = 2 × 2 + 2 × (−2) = 4 − 4 = 0.
When dot product is zero, vectors are perpendicular.
Therefore, the angle between them is 90°.

Solution:
For two-dimensional vectors:
A × B = (A_xB_y − A_yB_x) k
A × B = (3 × 4 − 2 × 1) k = (12 − 2) k = 10k.

Solution:
Area of parallelogram = |A × B|.
From the previous type of calculation:
A × B = (3 × 4 − 2 × 1) k = 10k.
So area = |10k| = 10 square units.

Solution:
Horizontal component:
u_x = u cos θ = 50 × 4/5 = 40 m s⁻¹
Vertical component:
u_y = u sin θ = 50 × 3/5 = 30 m s⁻¹.

Solution:
Initial speed:
u = √(u_x² + u_y²) = √(40² + 30²) = √2500 = 50 m s⁻¹.
Angle:
tan θ = u_y/u_x = 30/40 = 3/4.
Thus θ = tan⁻¹(3/4) = 36.9°.

Solution:
Time of flight is:
T = 2u sin θ / g
T = 2 × 20 × sin 30° / 10
T = 40 × 1/2 / 10 = 20/10 = 2 s.

Solution:
Maximum height is:
H = u² sin²θ / 2g
H = 20² × sin²30° / 20
H = 400 × (1/2)² / 20 = 400 × 1/4 / 20 = 100/20 = 5 m.

Solution:
Range is:
R = u² sin 2θ / g
R = 20² × sin 60° / 10
R = 400 × √3/2 / 10 = 20√3 m or approximately 34.6 m.

Solution:
For the same speed on horizontal ground, complementary angles give the same range.
Therefore, θ₁ + θ₂ = 90°.
Given θ₁ = 35°.
θ₂ = 90° − 35° = 55°.

Solution:
For horizontal projection, vertical initial velocity is zero.
h = 1/2 gt²
45 = 1/2 × 10 × t²
45 = 5t²
t² = 9
t = 3 s.

Solution:
From vertical motion, time of flight = 3 s.
Horizontal velocity remains constant.
Range = horizontal speed × time = 20 × 3 = 60 m.

Solution:
Centripetal acceleration is:
a_c = v²/r
a_c = 10²/5 = 100/5 = 20 m s⁻².

Solution:
Linear speed:
v = rω = 2 × 5 = 10 m s⁻¹.
Centripetal acceleration:
a_c = rω² = 2 × 5² = 2 × 25 = 50 m s⁻².