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Published by bm-1688, 2023-09-04 20:52:06

level up ch 2

SP015 Level UP 2 (Q)

LEVEL UP SP015 Ch 2 KMPk 2023 NAME: ……………………………………….………………………………. CLASS: ……………………… CHAPTER 2: KINEMATICS OF LINEAR MOTION 1. Sketch the graph of v-t and a-t for an object in linear motion. No. Graph of s - t Graph of v - t Graph of a - t (a) (b) (c) (d) v 0 t v 0 t a 0 t v 0 t s 0 t b a b s 0 t a b s 0 t a v 0 t a 0 t a 0 t s 0 t b a a 0 t


LEVEL UP SP015 Ch 2 KMPk 2023 2. Linear Motion Graph Analysis Choose a suitable word to complete the sentences in the table below. No. Graph of s - t Graph Analysis (a) Gradient of s - t graph represents velocity. Gradient of s - t graph is constant / increasing / decreasing from a to b. Hence, the velocity is constant / increasing / decreasing from a to b. The value of acceleration is ______________. (b) Gradient of s - t graph is positive / negative and the value is constant / increasing / decreasing from a to b. Hence, the velocity is constant / increasing / decreasing. The object is accelerating / decelerating. (c) Gradient of s - t graph is positive / negative and the value is constant / increasing / decreasing from a to b. Hence, the velocity is constant / increasing / decreasing. The object is accelerating / decelerating. (d) Gradient of s - t graph is positive / negative and the value is constant / increasing / decreasing from a to b. Hence, the velocity is constant / increasing / decreasing. The object is accelerating / decelerating. s 0 t b a b s 0 t a b s 0 t a s 0 t b a


LEVEL UP SP015 Ch 2 KMPk 2023 3. Free fall motion is a vertical motion of a body under the influence of _________________________ force only. 4. Complete the table on free fall motion below. Cases Analysis Case 1 An object is released from point A. u = Displacement of the object from initial position to the ground. s = –150 m Velocity of the object when it reaches point B. vB = Velocity of the object when it reaches point C. vC = –15 m s-1 Case 2 Object is thrown vertically upward. u = Displacement of the object from initial position to the ground. s = Velocity of the object when it reaches the maximum height. v = Velocity of the object when at point A. vA = 5 m s-1 20 m s-1 0.6 m A 5 m s-1 150 m 15 m s-1 B A C


LEVEL UP SP015 Ch 2 KMPk 2023 5. A marble is thrown vertically downward at 5 m s -1 from a height of 15 m. Calculate the (a) velocity of the object just before it hits the ground. (b) time taken by the object to reach the ground.


LEVEL UP SP015 Ch 2 KMPk 2023 6. Projectile motion is a motion that involves the resultant of two motions which is the ______________ motion (free fall) and the ______________ motion (x-component). 7. List down the kinematic equations for the types of motion below. Linear Motion Projectile Motion x - component y - component x x x x x v v u a t ; a 0 y y y y y v v u a t ; a g v u 2as 2 2 y y y y s s u t a t 2 2 1 s (u v)t 2 1 x x x x x x x s s u u t s u v t ( ) ( ) 2 1 2 1 8. Label parameters u, ux, sx, sy, v, vx, vy in the figure below.


LEVEL UP SP015 Ch 2 KMPk 2023 9. Complete the table on projectile motion below: Cases x - component y - component Case 1 ux = u cos 40 ux = 5 cos 40 ux = ______ m s-1 uy = When the object reaches point B. vx = vy = From initial point to the ground (A to C) sx = 100 m sy = Case 2 ux = uy = When the object reaches point B. vx = vy = –v sin 60 vy = –10 sin 60 vy = _______ m s-1 From initial point to the ground (A to C). sx = sy = Case 3 ux = uy = From initial point to the ground. sx = sy = When the object reaches the ground. vx = vy = ? u = 5 m s-1 40 34 100 m v = 4.62 m s-1 A B C B A u = 5 m s-1 150 m 60 120 m v = 10 m s-1 C 60 u = 5 m s-1 9.15 m 50 m v


LEVEL UP SP015 Ch 2 KMPk 2023 Cases x - component y - component Case 4 ux = uy = From initial point to the wall. sx = sy = When the object hits the wall. vx = .......... ...... s 1.5 3 t t s u t x x vy = ? -1 0.295m s 5.2 9.81(0.5) y y y y y v v v u a t Case 5 ux = uy = At maximum height vx = vy = From initial point to the wall. sx = sy = When the object hits the top of the wall. vx = ............... s 7.48 7.66 t t s u t x x vy = ? u = 10 m s -1 40 7.48 m 1.6 m 1.2 m u = 6 m s -1 60 1.5 m


LEVEL UP SP015 Ch 2 KMPk 2023 10. An object is launched upward at a velocity of 20 m s -1 in a direction making an angle of 25 with the horizontal. What is the (a) maximum height reached by the object? (b) total flight time (between launch and touching the ground) of the object? (c) horizontal range of the object? (d) magnitude of the velocity of the object just before it hits the ground?


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