Investigating Forces & Motion

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INI File | 1998-02-10 | 12KB | 417 lines

[question1] type:2 caption:\ Which one of the following is not an example of uniformly accelerated \ motion? correct:A swinging pendulum wrong1:A stone falling towards the ground wrong2:An astronaut jumping on the Moon wrong3:A car speeding up by 5.0 m/s each second feedback:\ The acceleration of a pendulum bob changes direction and size as it \ swings to and fro. It is the only one of the four examples that has a \ non-uniform acceleration. [question2] type:3 caption:\ The equations of uniformly accelerated motion apply to only one of \ these four objects. Which is it? correct:3g15d wrong1:3g15a wrong2:3g15b wrong3:3g15c feedback:\ The velocity-time graph of an object with uniform acceleration is a \ straight line. The only uniformly accelerating object of the four is \ (d). This is the only object to which the equations of uniformly \ accelerated motion can be applied. [question3] type:3 caption:\ Which one of the equations below is not an equation of uniformly \ accelerated motion? correct:3g21a wrong1:3g21b wrong2:3g21c wrong3:3g21d feedback:\ Equation (c) gives the kinetic (motion) energy of a moving mass. It is \ not one of the equations of uniformly accelerated motion. [question4] type:1 image:3g16 caption:\ By how much does the displacement of this object increase between \ t = 0.0 s and t = 4.0 s? correct:8.0 m wrong1:2.0 m wrong2:6.0 m wrong3:4.0 m feedback:\ The increase in displacement is equal to the area under the graph \ between t = 0.0 s and t = 4.0 s.\ Area = 4.0 x 2.0 = 8.0 m. [question5] type:1 image:3g17 caption:\ What is the increase in displacement of this object between\ t = 0.0 s and t = 2.0 s? correct:4.0 m wrong1:1.0 m wrong2:2.0 m wrong3:3.0 m feedback:\ The increase in displacement is equal to the area under the graph \ between t = 0.0 s and t = 2.0 s.\ Area = 1/2 x 2.0 x 4.0 = 4.0 m. [question6] type:1 image:3g18 caption:\ What is the increase in displacement of this object between\ t = 2.0 s and t = 3.0 s? correct:3.0 m wrong1:1.0 m wrong2:2.0 m wrong3:4.0 m feedback:\ The increase in displacement is equal to the area under the graph \ between t = 2.0 s and t = 3.0 s.\ Area = area of rectangle + area of triangle\ <center>= (1.0 x 2.0) + ½ x (1.0 x 2.0) = 3.0 m.</center> [question7] type:2 caption:\ A car travelling at 10 m/s starts to accelerate in a straight line \ with an acceleration 2.0 m/s2. How fast is it travelling \ 4.0 seconds later? correct:18 m/s wrong1:2.0 m/s wrong2:8.0 m/s wrong3:12 m/s feedback:\ Take the direction of travel as positive. Write down the knowns and \ unknowns in the problem.\ u = 10 m/s\ a = 2.0 m/s2\ t = 4.0 s\ v = ?\ Use v = u + at\ \ <center>= 10 + 2.0 x 4.0</center>\ <center>= 18 m/s.</center> [question8] type:2 caption:\ A car travelling at 10 m/s starts to accelerate in a straight line \ with an acceleration of 2.0 m/s2. How far does it travel in \ the next 4.0 seconds? correct:56 m wrong1:40 m wrong2:16 m wrong3:10 m feedback:\ Take the direction of travel as positive. Write down the knowns and \ unknowns in the problem.\ u= 10 m/s\ a = 2.0 m/s2\ t = 4.0 s\ s = ?\ Use s = ut + ½at2\ <center>= (10.0 x 4.0) + ½ x (2.0 x 4.02)</center>\ <center>= 56 m.</center> [question9] type:2 caption:\ A cyclist travelling at 8.0 m/s brakes. If the cyclist travels a \ further 8.0 m before coming to rest what was the average \ acceleration? correct:-4.0 m/s2 wrong1:-1.0 m/s2 wrong2:-2.0 m/s2 wrong3:-8.0 m/s2 feedback:\ Take the direction of travel as positive. Write down the knowns and \ unknowns in the problem.\ u = 8.0 m/s\ v = 0.0 m/s\ s = 8.0 m\ a = ?\ Use v2 = u2 + 2as\ \ 0.0 = 8.02 + 2a x 8.0\ 0 = 64 + 16a\ a = -64/16\ a = -4.0 m/s2. [question10] type:2 caption:\ A cyclist travelling at 8.0 m/s brakes but travels a further 8.0 m \ before coming to a complete rest. How long did it take for the bicycle \ to stop? correct:2.0 s wrong1:1.0 s wrong2:4.0 s wrong3:8.0 s feedback:\ Take the direction of travel as positive. Write down the knowns and \ unknowns in the problem.\ u = 8.0 m/s\ v = 0.0 m/s\ s = 8.0 m\ t = ?\ Use <img src="sa3q10a" align=center>\ 8.0 = ½ (8.0 + 0.0) t\ \ 16 = 8.0t\ \ t = 2.0 s. [question11] type:2 caption:\ A pebble is dropped from a cliff 20 m above the beach. How long is it \ before the pebble hits the sand? (Assume that the acceleration due to \ gravity is g = 10 m/s2.) correct:2.0 s wrong1:1.0 s wrong2:3.0 s wrong3:4.0 s feedback:\ Take 'down' as positive. Write down the knowns and unknowns in the \ problem.\ u = 0.0 m/s\ s = 20 m\ a = 10 m/s2\ t = ?\ Use s = ut + ½at2\ 20 = 0.0 + ½ x 10 x t2\ t2 = 20/5\ t2 = 4.0\ t = 2.0 s. [question12] type:2 caption:\ A pebble is dropped from a cliff 20 m above the beach. How fast is the \ pebble travelling when it hits the sand? (Assume that air resistance \ is negligible and the acceleration due to gravity, g = 10 \ m/s2.) correct:20 m/s wrong1:2.0 m/s wrong2:5.0 m/s wrong3:10 m/s feedback:\ Take 'down' as positive. Write down the knowns and unknowns in the \ problem.\ u = 0.0 m/s\ s = 20 m\ a = 10 m/s2\ v = ?\ Use v2 = u2 + 2as\ \ v2 = 0.02 + 2.0 x 10 x 20\ v2 = 20 x 20\ v = 20 m/s. [question13] type:2 caption:\ A juggler throws a ball vertically into the air and catches it at the \ same level 2.0 s later. How high did the ball rise? (Assume that air \ resistance is negligible and the acceleration due to gravity, g \ = 10 m/s2.) correct:5.0 m wrong1:2.0 m wrong2:10 m wrong3:20 m feedback:\ Take 'down' as positive. The ball's velocity when it reaches its \ highest point will be zero. Consider the second half of the motion \ only. The initial velocity is zero. The time to fall back from the \ highest point is half the total time of 2.0 s. Write down the knowns \ and unknowns in the problem.\ u = 0.0 m/s\ t = 1.0 s\ a = 10 m/s2\ s = ?\ Use s = ut+ ½at2\ s = 0.0 + ½ x 10 x 1.02\ s = 5.0 m. [question14] type:2 caption:\ A juggler throws a ball vertically into the air and catches it at the \ same level 2.0 s later. How fast was the ball travelling when it left \ the juggler's hand? (Assume that air resistance is negligible the \ acceleration due to gravity, g = 10 m/s2.) correct:10 m/s wrong1:2.0 m/s wrong2:5.0 m/s wrong3:20 m/s feedback:\ Take 'up' as positive. Consider the first half of the motion only. The \ ball's velocity when it reaches its highest point will be zero. The \ time to reach the highest point is half the total time of 2.0 s. Write \ down the knowns and unknowns in the problem.\ v = 0.0 m/s\ t = 1.0 s\ a = -10 m/s2\ u = ?\ Use v = u + at\ \ 0.0 = u - 10 x 1.0\ u = 10 m/s. [question15] type:2 caption:\ A frog leaps vertically into the air at a speed of 2.0 m/s. How high \ does it jump? (Assume that air resistance is negligible and the \ acceleration due to gravity, g = 10 m/s2.) correct:0.2 m wrong1:0.5 m wrong2:1.0 m wrong3:2.0 m feedback:\ Take 'up' as positive. The frog's velocity when it reaches its highest \ point will be zero. Write down the knowns and unknowns in the \ problem.\ u = 2.0 m/s\ v = 0.0 m/s\ a = -10 m/s2\ s = ?\ Use v2 = u2 + 2as\ \ 0 = 22 - 2 x 10 x s\ \ 20 s = 4.0\ s = 4.0/20\ s = 0.2 m. [question16] type:2 caption:\ A frog leaps vertically into the air at a speed of 2.0 m/s. How long \ is it in the air? (Assume that air resistance is negligible and the \ acceleration due to gravity, g = 10 m/s2.) correct:0.4 s wrong1:0.2 s wrong2:1.0 s wrong3:2.4 s feedback:\ Take 'up' as positive. The frog's velocity when it reaches its highest \ point will be zero. The time to reach the highest point will be half \ the total time in the air. Write down the knowns and unknowns in the \ problem.\ u = 2.0 m/s\ v = 0.0 m/s\ a = -10 m/s2\ t = ?\ Use v = u + at\ \ 0.0 = 2 - 10t\ \ 10t = 2\ t = 2/10\ t = 0.2 s\ Therefore total time = 2.0 x 0.2 = 0.4 s. [question17] type:2 caption:\ A space craft must decelerate from a speed 10 000 m/s to 2 000 m/s as \ it approaches a moon. If the maximum acceleration that the bodies of \ the astronauts can withstand is 100 m/s2 how long will the \ deceleration take? correct:80 s wrong1:1 000 s wrong2:800 s wrong3:100 s feedback:\ Take the direction of motion as positive. Write down the knowns and \ unknowns in the problem.\ u = 10 000 m/s\ v = 2 000 m/s\ a = -100 m/s2\ t = ?\ Use v = u + at\ \ 2 000 = 10 000 - 100t\ \ 100t = 8 000\ t = 8 000/100\ t = 80 s. [question18] type:2 caption:\ A space craft must decelerate from a speed 10 000 m/s to 2 000 m/s as \ it approaches a moon. If the maximum acceleration that the bodies of \ the astronauts can withstand is 100 m/s2 how far from the \ moon must the deceleration start? correct:480 km wrong1:480 m wrong2:4 800 m wrong3:4 800 km feedback:\ Take the direction of motion as positive. Write down the knowns and \ unknowns in the problem.\ u = 10 000 m/s\ v = 2 000 m/s\ a = -100 m/s2\ s = ?\ Use v2 = u2 + 2as\ \ 2 0002 = 10 0002 - 2 x 100t\ \ 200t = 100 000 000 - 4 000 000\ t = 96 000 000/200\ t = 480 000 m\ t = 480 km. [question19] type:1 image:3g19 caption:\ This graph shows how the velocity of an object undergoing a uniform \ acceleration changes with time. What is the object's acceleration? correct:1.0 m/s2 wrong1:2.0 m/s2 wrong2:5.0 m/s2 wrong3:10 m/s2 feedback:\ The gradient of a velocity-time graph is the acceleration.\ Gradient = 1.0 m/s2. [question20] type:1 image:3g20 caption:\ This graph shows how the velocity of an object undergoing a uniform \ acceleration changes with time. By how much does the object's \ displacement increase between t = 1.0 s and t = 3.0 s? correct:6.0 m wrong1:1.0 m wrong2:2.0 m wrong3:5.0 m feedback:\ The area under a velocity-time graph is the increase in \ displacement.\ area between t = 1.0 s and t = 3.0 s is equal to the area of the \ rectangle plus the area of the triangle.\ displacement = 2.0 x 2.0 + 1/2 x 2.0 x 2.0\ displacement = 6.0 m.