Investigating Forces & Motion

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INI File | 1998-02-10 | 13KB | 414 lines

[question1] type:2 caption:\ What is the momentum of an object with mass, m, travelling at a \ velocity, v? correct:mv wrong1:m/v wrong2:v/m wrong3:mv2 feedback:\ Momentum equals mass multiplied by velocity = mv.\ [question2] type:2 caption:What are the correct units of momentum? correct:kg m/s wrong1:kg wrong2:m/s2 wrong3:J feedback:\ The units of momentum are those of mass multiplied by velocity = kg \ m/s. [question3] type:1 image:7g14 caption:\ An athlete has a mass of 80 kg. What is the athlete's momentum when \ running at a speed of 10 m/s? correct:800 kg m/s wrong1:8.0 kg m/s wrong2:70 kg m/s wrong3:80 kg m/s feedback:\ Momentum = mass x velocity\ <center>= 80 x 10</center>\ <center>= 800 kg m/s.</center> [question4] type:1 image:7g15 caption:\ A bird has a mass of 0.1 kg. What is its momentum when it is flying at \ a speed of 5.0 m/s? correct:0.5 kg m/s wrong1:0.1 kg m/s wrong2:5.0 kg m/s wrong3:50 kg m/s feedback:\ Momentum = mass x velocity\ <center>= 0.1 x 5.0</center>\ <center>= 0.5 kg m/s.</center> [question5] type:2 caption:\ A cyclist travelling at 15 m/s has a momentum of 1 200 kg m/s. What is \ the total mass of the cyclist and the bicycle? correct:80 kg wrong1:1.25 kg wrong2:18 kg wrong3:18 000 kg feedback:\ Momentum = mass x velocity\ Therefore,\ <img src="sa7q5a" align=center>\ <img src="sa7q5b" align=center>\ <center>= 80 kg.</center> [question6] type:2 caption:\ A car with mass 750 kg has a momentum of 15 000 kg m/s. How fast is it \ travelling? correct:20 m/s wrong1:15 000 m/s wrong2:1 125 m/s wrong3:0.05 m/s feedback:\ Momentum = mass x velocity\ Therefore,\ <img src="sa7q6a" align=center>\ <img src="sa7q6b" align=center>\ <center>= 20 m/s.</center> [question7] type:2 caption:\ The impulse of a force, F, on an object is the product of the \ force and the length of time, t, for which the force acts. To \ which one of the following is the impulse equal? correct:The change in the momentum of the object. wrong1:The change in the velocity of the object. wrong2:The change in the energy of the object. wrong3:The change in the position of the object. feedback:The impulse is equal to the change in momentum. [question8] type:2 caption:What are the correct units of impulse? correct:kg m/s wrong1:kg wrong2:m/s2 wrong3:J feedback:\ Because the impulse equals the change in momentum, its units are the \ same as those of momentum: kg m/s. [question9] type:2 caption:\ A force of 10 N accelerates a 2.0 kg mass for 4.0 s in a straight \ line. By how much does the speed of the object change? correct:20 m/s wrong1:2.0 m/s wrong2:8.0 m/s wrong3:10 m/s feedback:\ Impulse = Ft\ \ <center>= 10 x 4.0</center>\ <center>= 40 kg m/s</center>\ This is the object's momentum change.\ momentum = mass x velocity\ Therefore,\ <img src="sa7q9a" align=center>\ <img src="sa7q9b" align=center>\ <center>= 20 m/s.</center> [question10] type:2 caption:\ A 2.0 kg ball of soft clay thrown against a wall comes to rest in 0.05 \ s. If the speed of the ball just before it hits the wall is 10 m/s, \ what is the average impact force on the wall as the clay ball strikes \ it? correct:400 N wrong1:2.0 N wrong2:0.10 N wrong3:40 N feedback:\ Use the fact that impulse = momentum change to find the force.\ initial momentum = mass x velocity\ <center>= 2.0 x 10</center>\ <center>= 20 kg m/s</center>\ final momentum = 0.0\ momentum change = final momentum - initial momentum\ <center>= 0.0 - 20</center>\ <center>= -20 kg m/s</center>\ impulse = Ft = momentum change\ Therefore,\ <img src="sa7q10a" align=center>\ <img src="sa7q10b" align=center>\ <center>= -400 N</center>\ This is the force on the ball. The force on the wall is equal and \ opposite to this force (Newton's third law of motion), and is \ therefore 400 N. [question11] type:2 caption:\ The average force of a cricket bat on a ball is 800 N. If the momentum \ of the ball changes by 2.0 kg m/s, for how long is the ball in contact \ with the bat? correct:0.0025 s wrong1:0.10 s wrong2:1.6 s wrong3:2.0 s feedback:\ Impulse = force x time = change in momentum\ Therefore,\ <img src="sa7q11a" align=center>\ <img src="sa7q11b" align=center>\ <center>= 0.0025 s.</center> [question12] type:1 image:7g16 caption:\ The diagram shows a ball bouncing off a wall. The mass of the ball is \ 0.1 kg. What is the ball's momentum change? correct:-0.6 kg m/s wrong1:zero wrong2:-3.0 kg m/s wrong3:+3.0 kg m/s feedback:\ Momentum = mass x velocity\ Momentum change = final momentum - initial momentum\ <center>= (-3.0 x -0.1) - (3.0 x 0.1)</center>\ <center>= -0.3 - 0.3</center>\ <center>= -0.6 kg m/s.</center> [question13] type:1 image:7g17 caption:\ The diagram shows a 3.0 kg ball of clay dropped onto a hard floor. If \ the speed of the ball just before it hits the floor is 5.0 m/s, what \ is its momentum change as it comes to rest? correct:-15 kg m/s wrong1:zero wrong2:-1.7 kg m/s wrong3:30 kg m/s feedback:\ Momentum = mass x velocity\ Momentum change = final momentum - initial momentum\ <center>= 0.0 - (3.0 x 5.0)</center>\ <center>= -15 kg m/s.</center> [question14] type:2 caption:\ Which one of the following suggestions would not make a car safer in \ an accident ? correct:A. Make the car stronger and stiffer so that its body crumples less. wrong1:B. Build in thick bumpers that crush upon impact. wrong2:\ C. Make sure that the seat belts hold the driver and passengers \ securely, so that their momentum changes at the same rate as the car's \ momentum. wrong3:D. Fit air bags to absorb the front-seat passenger's forward momentum. feedback:\ Suggestion (A) would not make a car safer. If cars were stiffer and \ stronger so that they crumpled less in accidents, then they would come \ to rest in shorter times as they crashed. Because the change in \ momentum is equal to force x time, reducing the time would increase \ the force. The forces on passengers strapped into the car would \ therefore also be greater, so they would be injured more severely. [question15] type:1 image:7g18 caption:\ The diagram shows a 0.05 kg arrow being shot into an apple hanging \ from a string. The speed of the arrow just before impact is 30 m/s. If \ the mass of the apple is 0.1 kg and the arrow sticks into the apple, \ what is the speed of the apple and arrow immediately after the \ impact? correct:10 m/s wrong1:1.0 m/s wrong2:25 m/s wrong3:100 m/s feedback:\ The principle of the conservation of momentum applies to this \ impact.\ initial momentum = final momentum\ 0.05 x 30 = (0.05 + 0.1) x V\ \ (In the above equation, v is the speed of the arrow and apple \ immediately after the impact).\ Thus,\ <img src="sa7q15a" align=center>\ <center>= 10 m/s.</center> [question16] type:1 image:7g19 caption:\ The two railway trucks shown are travelling on a smooth track. They \ collide and couple together. What is their speed after the \ collision? correct:zero wrong1:2.0 m/s to the left wrong2:4.0 m/s to the right wrong3:6.0 m/s to the right feedback:\ The principle of the conservation of momentum applies to this \ impact.\ initial momentum = final momentum\ 1 000 x 4.0 + 2 000 x -2.0 = (1 000 + 2 000) x V\ \ (In the above equation, V is the speed of the trucks \ immediately after the impact.)\ So,\ <img src="sa7q16a" align=center>\ <center>= zero.</center> [question17] type:1 image:7g20 caption:\ Two ice skaters standing on a smooth rink push each other apart. The \ mass of one skater is 30 kg, the mass of the other is 60 kg. If the \ lighter skater moves off with a velocity of 2.0 m/s, what is the \ velocity of the heavier skater? correct:-1.0 m/s wrong1:-2.0 m/s wrong2:-4.0 m/s wrong3:-8.0 m/s feedback:\ The principle of the conservation of momentum applies to this \ situation. The skaters' initial momentum is zero, so their final total \ momentum must also be zero.\ Let the final velocity of the heavier skater be v.\ momentum = mass x velocity\ initial momentum = final momentum\ 0.0 = 30 x 2.0 + 60 x v\ \ Therefore,\ 60v = -60\ <img src="sa7q17a" align=center>\ <center>= -1.0 m/s.</center> [question18] type:1 image:7g21 caption:\ A 1 000 kg car and a 3 000 kg lorry travelling in opposite directions \ collide head-on. They both come to rest at the point of impact. If the \ car is travelling at a velocity of 30 m/s before the crash, what is \ the lorry's velocity? correct:-10 m/s wrong1:-90 m/s wrong2:-30 m/s wrong3:-9.0 m/s feedback:\ Assuming there are no external forces then the principle of the \ conservation of momentum applies to this impact. The vehicles come to \ rest at the point of impact, so their final total momentum is \ zero.\ Let U be the initial velocity of the lorry.\ initial momentum = final momentum\ 1 000 x 30 + 3 000 x U = 0.0\ Thus:\ 3 000U = -30 000\ <img src="sa7q18a" align=center>\ <center>= -10 m/s.</center> [question19] type:1 image:7g22 caption:\ Two masses collide and rebound with the velocities shown. Which of the \ following equations is a correct statement of the principle of the \ conservation of momentum? correct:\ m<TT>1</TT>u<TT>1</T \ > + \ m<TT>2</TT>u<TT>2</TT>< \ > = \ m<TT>1</TT>v<TT>1</TT> \ + \ m<TT>2</TT>v<TT>2</TT> wrong1:\ m<TT>1</TT>u<TT>1</T \ > - \ m<TT>2</TT>u<TT>2</T \ > = \ m<TT>1</TT>v<TT>1</TT>< \ > - \ m<TT>2</TT>v<TT>2</TT> wrong2:\ m<TT>1</TT>u<TT>1</T \ > + m<TT>2</TT>v1 \ = \ m<TT>1</TT>u2 + \ m<TT>2</TT>v<TT>2</TT> wrong3:\ m<TT>1</TT>u<TT>1</T \ > x \ m<TT>2</TT>u<TT>2</TT>< \ > = \ m<TT>1</TT>v<TT>1</TT>< \ > x \ m<TT>2</TT>v<TT>2</TT> feedback:\ The principle of the conservation of momentum states that, if there \ are no external forces acting upon a system, then initial momentum = \ final momentum.\ In this case, the initial momentum is \ m<TT>1</TT>u<TT>1</T \ > + \ m<TT>2</TT>u<TT>2</TT> \ and the final momentum is \ m<TT>1</TT>v<TT>1</TT>< \ > + \ m<TT>2</TT>v<TT>2. \ Equation (b) is therefore correct:\ </TT>\ m<TT>1</TT>u<TT>1</T \ > + \ m<TT>2</TT>u<TT>2</TT>< \ > = \ m<TT>1</TT>v<TT>1</TT>< \ > + m<TT>2</TT>v<TT>2\ </TT>\ <TT> [question20] type:1 image:7g23 caption:\ A 100 kg astronaut pushes a 500 kg satellite in space. If the \ satellite moves off at a velocity of 0.1 m/s, what is the astronaut's \ velocity after they part company? correct:-0.5 m/s wrong1:-0.1 m/s wrong2:0.05 m/s wrong3:0.02 m/s feedback:\ The principle of the conservation of momentum applies to this \ situation. The combined momentum of the satellite and the astronaut is \ initially zero, so their final total momentum must also be zero.\ Let the final velocity of the astronaut be v.\ momentum = mass x velocity\ initial momentum = final momentum\ 0 = 100v + 500 x 0.1\ Therefore,\ 100v = -50\ <img src="sa7q20a" align=center>\ <center>= -0.5 m/s.</center>