Physics 111 Past Paper PDF

Phy 111 KINEMATICS; EQUATIONS OF MOTION AND PROJECTILES OBJECTIVES To understand Kinematic concept Solve problems involving Equations of motion Understand the concept of projectiles Solve Problems involving trajectories  Kinematics is the study of the motion of objects without refere...

Phy 111 KINEMATICS; EQUATIONS OF MOTION AND PROJECTILES OBJECTIVES To understand Kinematic concept Solve problems involving Equations of motion Understand the concept of projectiles Solve Problems involving trajectories  Kinematics is the study of the motion of objects without reference to the forces that caused the motion. It is also the study of motion with respect to space and time. Equations of motion There are linear equations of motion Horizontal Direction Motion under gravity 𝑉 = 𝑢 + 𝑎𝑡 𝑉 = 𝑢 ± 𝑔𝑡 𝑉 2 = 𝑢2 + 2𝑎𝑆 𝑉 2 = 𝑢2 ± 2𝑔ℎ 1 2 1 2 𝑆 = 𝑢𝑡 + 𝑎𝑡 ℎ = 𝑢𝑡 ± 𝑔𝑡 2 2 𝑈+𝑉 𝑆 = 𝑡 2 where v = Final velocity (m/s), u = initial velocity (m/s), S = distance covered in the horizontal direction (m), h = height = distance covered in vertical direction (m), t = time spent during the motion (s).  EXAMPLES (1) A Body at rest was accelerated, and it traveled for 5 minutes. The constant acceleration provided to the body was 2 m/s2. Find the final velocity the object had before coming to a stop. solution u = 0 m/s, a = 2 m/s2 , t = 5 min = 5 x 60 = 300s v = u + at = 0 + 2 x 300 = 0 + 600 = 600 m/s Hence, the final velocity is 600 m/s (2) Find the final velocity of an object that traveled a distance of 500 m with an initial speed of 2 m/s and acceleration of the object was 0.5m/s2. Solution 𝑉 2 = 𝑢2 + 2𝑎𝑆 , u = 2 m/s, a = 0.5 m/s2 , s = 500 m, v = ? v2 = (2)2 + 2 x 0.5 x 500 = 4 + 500 = 504 𝑉 = 504 = 22.45 , Hence, the Final velocity is 22.45 m/s (3) A ball is dropped from a certain height. It took the ball 15 seconds to reach the ground. What is the height at which the ball was initially? [Take g=10m/s2] 1 ℎ = 𝑢𝑡 + 𝑔𝑡 2 , u = 0 m/s, t = 15 s, g = 10 m/s2 2 h = 0 x 15 + 0.5 x 10 x 152 = 5 x 225 = 1125 The height is 1125 m (3) The velocity acquired by an object moving with uniform acceleration is 60 m/s in 3 seconds and 120 m/s in 6 seconds. Find the initial velocity. (3) The velocity acquired by an object moving with uniform acceleration is 60 m/s in 3 seconds and 120 m/s in 6 seconds. Find the initial velocity. Solution Using the equation, v = u + at, we have; 60 = u + 3a.........(1) 120 = u + 6a..........(2) making u, subject of the formular in equ (1), we have u = 60 - 3a... (3), subst. equ (3) into equ (2), we have 120 = 60 - 3a 120 = 60 + 3a 3a = 120 - 60 3a = 60, a = 20 m/s2 , subst, a into any of the equations, e.g equ (1) 60 = u + 3 (20) 60 = u + 60 u = 60 - 60 = 0 u = 0 m/s , Therefore, the initial velocity, u = 0 m/s (4) A ball is thrown upwards with a velocity of 55 m/s. Find the velocity after 4s. Also find out the maximum height attained by the ball. (5) A stone is thrown vertically up with a velocity of 10 m/s. Find (a) The maximum height reached by the stone. (b) The time taken to reach the maximum height. (c) The velocity with which it touches the ground. (d) The time taken to reach the ground, g may be taken as 9.8 m/s2 (6) A ball thrown vertically upwards from the top of a building of height 80 m returns to the earth after 10 s. What is the velocity of the projection? Also g may be taken as 10 m/s2. (7) An object moving with an initial velocity of 15 m/s, has a uniform acceleration of 3 m/s2. Find the distance travelled by the object in the 10th second of its motion. (8) An object is dropped from a height of 80 metres. What will be the distance travelled by it during the last second of its fall? Also, take g = 10 m/s2  A projectile is any object thrown into space upon which the only acting force is gravity. The primary force acting on a projectile is gravity. This doesn’t necessarily mean that other forces do not act on it, just that their effect is minimal compared to gravity. The path followed by a projectile is known as a trajectory. A baseball batted or thrown is an example of a projectile. In a Projectile Motion, there are two simultaneous independent rectilinear motions: Along the x-axis: uniform velocity, responsible for the horizontal (forward) motion of the particle. Along the y-axis: uniform acceleration, responsible for the vertical (downwards) motion of the particle.  Projectiles equations 𝒖𝑺𝒊𝒏𝜽 𝒕 = where t is the time it takes for body to reach the 𝒈 maximum height 𝟐𝒖𝑺𝒊𝒏𝜽 𝑻 = Where T is the total time of flight 𝒈 𝒖𝟐 𝑺𝒊𝒏𝟐 𝜽 𝑯 = H is the maximum height (vertical distance) 𝟐𝒈 attained. 𝒖𝟐 𝑺𝒊𝒏 𝟐𝜽 𝑹 = R is the range (horizontal distance) covered. 𝒈 (9) An object of mass 2000 g covers a maximum vertical distance of 6 m when it is projected at an angle of 45° from the ground. Calculate the velocity with which it was thrown. Take g = 10 m/s2. Solution 𝑢2 𝑆𝑖𝑛2 𝜃 𝐻 = , θ = 45⁰, H = 6 m, g = 10 m/s, u = ?, 2𝑔 𝒖𝟐 𝑺𝒊𝒏 𝟒𝟓 ×𝑺𝒊𝒏𝟒𝟓 𝒖𝟐 𝟎.𝟓 𝟔 = 𝟐 ×𝟏𝟎 = 𝟐𝟎 , 20 x 6 = u2 x 0.5 𝟐 𝟐𝟎 ×𝟔 𝒖 = 𝟎.𝟓 = 𝟐𝟒𝟎 , 𝒖 = 𝟐𝟒𝟎 = 𝟏𝟓. 𝟒𝟗 the initial velocity is therefore , u = 15.49 m/s (10) An object is launched at a velocity of 20 m/s in a direction making an angle of 25° upward with the horizontal. a) What is the maximum height reached by the object? b) What is the total flight time (between launch and touching the ground) of the object? c) What is the horizontal range (maximum x above ground) of the object? d) What is the magnitude of the velocity of the object just before it hits the ground? (11) A ball kicked from ground level at an initial velocity of 60 m/s and an angle θ with ground reaches a horizontal distance of 200 meters. a) What is the size of angle θ? b) What is time of flight of the ball? (12) A bullet shot hits the ground 4 km away, which was at an angle of 30° with the plane. What should be the projection angle to hit a target 6 km away? Neglect air resistance and assume fixed muzzle speed (13) A soccer ball is kicked with an initial speed of 25 m/s. What is the maximum horizontal range it can achieve? (Assume g = 9.8 m/s²) (14) A stone is thrown from the top of a 30-meter-high cliff at an angle of 60 degrees to the horizontal with an initial speed of 20 m/s. What is its total time of flight? (Assume g = 9.8 m/s²)  (15)Question: A ball is kicked at an angle of 60 degrees to the horizontal with an initial speed of 25 m/s. What is its vertical velocity at its highest point? (Assume g = 9.8 m/s²) THANK YOU SO MUCH FOR LEARNING.