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INI File | 1998-02-10 | 12.7 KB | 414 lines |
- [question1]
- type:2
- caption:\
- What is the momentum of an object with mass, <I>m</I>, travelling at a \
- velocity, <I>v</I>?<p>
- correct:<I>mv</I>
- wrong1:<I>m</I>/<I>v</I>
- wrong2:<I>v</I>/<I>m</I>
- wrong3:<I>mv</I><sup>2</sup>
- feedback:\
- Momentum equals mass multiplied by velocity = <I>mv.<p>\
- </I><p>
-
- [question2]
- type:2
- caption:What are the correct units of momentum?<p>
- correct:kg m/s
- wrong1:kg
- wrong2:m/s<sup>2</sup>
- wrong3:J
- feedback:\
- The units of momentum are those of mass multiplied by velocity = kg \
- m/s.<p>
-
- [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?<p>
- 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<p>\
- <center>= 80 x 10</center><p>\
- <center>= 800 kg m/s.</center><p>
-
- [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?<p>
- 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<p>\
- <center>= 0.1 x 5.0</center><p>\
- <center>= 0.5 kg m/s.</center><p>
-
- [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?<p>
- correct:80 kg
- wrong1:1.25 kg
- wrong2:18 kg
- wrong3:18 000 kg
- feedback:\
- Momentum = mass x velocity<p>\
- Therefore,<p>\
- <img src="sa7q5a" align=center><p>\
- <img src="sa7q5b" align=center><p>\
- <center>= 80 kg.</center><p>
-
- [question6]
- type:2
- caption:\
- A car with mass 750 kg has a momentum of 15 000 kg m/s. How fast is it \
- travelling?<p>
- correct:20 m/s
- wrong1:15 000 m/s
- wrong2:1 125 m/s
- wrong3:0.05 m/s
- feedback:\
- Momentum = mass x velocity<p>\
- Therefore,<p>\
- <img src="sa7q6a" align=center><p>\
- <img src="sa7q6b" align=center><p>\
- <center>= 20 m/s.</center><p>
-
- [question7]
- type:2
- caption:\
- The impulse of a force, <I>F</I>, on an object is the product of the \
- force and the length of time, <I>t</I>, for which the force acts. To \
- which one of the following is the impulse equal?<p>
- 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.<p>
-
- [question8]
- type:2
- caption:What are the correct units of impulse?<p>
- correct:kg m/s
- wrong1:kg
- wrong2:m/s<sup>2</sup>
- wrong3:J
- feedback:\
- Because the impulse equals the change in momentum, its units are the \
- same as those of momentum: kg m/s.<p>
-
- [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?<p>
- correct:20 m/s
- wrong1:2.0 m/s
- wrong2:8.0 m/s
- wrong3:10 m/s
- feedback:\
- Impulse = <I>Ft<p>\
- </I><p>\
- <center>= 10 x 4.0</center><p>\
- <center>= 40 kg m/s</center><p>\
- This is the object's momentum change.<p>\
- momentum = mass x velocity<p>\
- Therefore,<p>\
- <img src="sa7q9a" align=center><p>\
- <img src="sa7q9b" align=center><p>\
- <center>= 20 m/s.</center><p>
-
- [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?<p>
- 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.<p>\
- initial momentum = mass x velocity<p>\
- <center>= 2.0 x 10</center><p>\
- <center>= 20 kg m/s</center><p>\
- final momentum = 0.0<p>\
- momentum change = final momentum - initial momentum<p>\
- <center>= 0.0 - 20</center><p>\
- <center>= -20 kg m/s</center><p>\
- impulse = <I>Ft</I> = momentum change<p>\
- Therefore,<p>\
- <img src="sa7q10a" align=center><p>\
- <img src="sa7q10b" align=center><p>\
- <center>= -400 N</center><p>\
- 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.<p>
-
- [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?<p>
- correct:0.0025 s
- wrong1:0.10 s
- wrong2:1.6 s
- wrong3:2.0 s
- feedback:\
- Impulse = force x time = change in momentum<p>\
- Therefore,<p>\
- <img src="sa7q11a" align=center><p>\
- <img src="sa7q11b" align=center><p>\
- <center>= 0.0025 s.</center><p>
-
- [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?<p>
- 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<p>\
- Momentum change = final momentum - initial momentum<p>\
- <center>= (-3.0 x -0.1) - (3.0 x 0.1)</center><p>\
- <center>= -0.3 - 0.3</center><p>\
- <center>= -0.6 kg m/s.</center><p>
-
- [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?<p>
- correct:-15 kg m/s
- wrong1:zero
- wrong2:-1.7 kg m/s
- wrong3:30 kg m/s
- feedback:\
- Momentum = mass x velocity<p>\
- Momentum change = final momentum - initial momentum<p>\
- <center>= 0.0 - (3.0 x 5.0)</center><p>\
- <center>= -15 kg m/s.</center><p>
-
- [question14]
- type:2
- caption:\
- Which one of the following suggestions would not make a car safer in \
- an accident ?<p>
- 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.<p>
-
- [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?<p>
- 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.<p>\
- initial momentum = final momentum<p>\
- 0.05 x 30 = (0.05 + 0.1) x <I>V<p>\
- </I><p>\
- (In the above equation, <I>v</I> is the speed of the arrow and apple \
- immediately after the impact).<p>\
- Thus,<p>\
- <img src="sa7q15a" align=center><p>\
- <center>= 10 m/s.</center><p>
-
- [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?<p>
- 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.<p>\
- initial momentum = final momentum<p>\
- 1 000 x 4.0 + 2 000 x -2.0 = (1 000 + 2 000) x <I>V<p>\
- </I><p>\
- (In the above equation, <I>V</I> is the speed of the trucks \
- immediately after the impact.)<p>\
- So,<p>\
- <img src="sa7q16a" align=center><p>\
- <center>= zero.</center><p>
-
- [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?<p>
- 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.<p>\
- Let the final velocity of the heavier skater be <I>v</I>.<p>\
- momentum = mass x velocity<p>\
- initial momentum = final momentum<p>\
- 0.0 = 30 x 2.0 + 60 x <I>v<p>\
- </I><p>\
- Therefore,<p>\
- 60<I>v</I> = -60<p>\
- <img src="sa7q17a" align=center><p>\
- <center>= -1.0 m/s.</center><p>
-
- [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?<p>
- 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.<p>\
- Let <I>U</I> be the initial velocity of the lorry.<p>\
- initial momentum = final momentum<p>\
- 1 000 x 30 + 3 000 x <I>U</I> = 0.0<p>\
- Thus:<p>\
- 3 000<I>U</I> = -30 000<p>\
- <img src="sa7q18a" align=center><p>\
- <center>= -10 m/s.</center><p>
-
- [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?<p>
- correct:\
- <I>m</I><TT><I><SUB>1</SUB></I></TT><I>u</I><TT><I><SUB>1</SUB></I></T \
- > +<I> \
- m</I><TT><I><SUB>2</SUB></I></TT><I>u</I><TT><I><SUB>2</SUB></I></TT>< \
- > </I>=<I> \
- m</I><TT><I><SUB>1</SUB></I></TT><I>v</I><TT><I><SUB>1</SUB></I></TT> \
- +<I> \
- m</I><TT><I><SUB>2</SUB></I></TT><I>v</I><TT><I><SUB>2</SUB></I></TT>
- wrong1:\
- <I>m</I><TT><I><SUB>1</SUB></I></TT><I>u</I><TT><I><SUB>1</SUB></I></T \
- ><I> </I>- \
- <I>m</I><TT><I><SUB>2</SUB></I></TT><I>u</I><TT><I><SUB>2</SUB></I></T \
- > =<I> \
- m</I><TT><I><SUB>1</SUB></I></TT><I>v</I><TT><I><SUB>1</SUB></I></TT>< \
- > </I>-<I> \
- m</I><TT><I><SUB>2</SUB></I></TT><I>v</I><TT><I><SUB>2</SUB></I></TT>
- wrong2:\
- <I>m</I><TT><I><SUB>1</SUB></I></TT><I>u</I><TT><I><SUB>1</SUB></I></T \
- ><I> </I>+<I> m</I><TT><I><SUB>2</SUB></I></TT><I>v<I><SUB>1</SUB> \
- </I></I>=<I> \
- m</I><TT><I><SUB>1</SUB></I></TT><I>u<I><SUB>2</SUB></I></I> +<I> \
- m</I><TT><I><SUB>2</SUB></I></TT><I>v</I><TT><I><SUB>2</SUB></I></TT>
- wrong3:\
- <I>m</I><TT><I><SUB>1</SUB></I></TT><I>u</I><TT><I><SUB>1</SUB></I></T \
- > x<I> \
- m</I><TT><I><SUB>2</SUB></I></TT><I>u</I><TT><I><SUB>2</SUB></I></TT>< \
- > </I>=<I> \
- m</I><TT><I><SUB>1</SUB></I></TT><I>v</I><TT><I><SUB>1</SUB></I></TT>< \
- > </I>x<I> \
- m</I><TT><I><SUB>2</SUB></I></TT><I>v</I><TT><I><SUB>2</SUB></I></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.<p>\
- In this case, the initial momentum is \
- <I>m</I><TT><I><SUB>1</SUB></I></TT><I>u</I><TT><I><SUB>1</SUB></I></T \
- ><I> </I>+<I> \
- m</I><TT><I><SUB>2</SUB></I></TT><I>u</I><TT><I><SUB>2</SUB></I></TT> \
- and the final momentum is<I> \
- m</I><TT><I><SUB>1</SUB></I></TT><I>v</I><TT><I><SUB>1</SUB></I></TT>< \
- > </I>+<I> \
- m</I><TT><I><SUB>2</SUB></I></TT><I>v</I><TT><I><SUB>2</SUB></I><SUB>. \
- </SUB>Equation (b) is therefore correct:<p>\
- </TT><p>\
- <I>m</I><TT><I><SUB>1</SUB></I></TT><I>u</I><TT><I><SUB>1</SUB></I></T \
- ><I> </I>+<I> \
- m</I><TT><I><SUB>2</SUB></I></TT><I>u</I><TT><I><SUB>2</SUB></I></TT>< \
- > </I>=<I> \
- m</I><TT><I><SUB>1</SUB></I></TT><I>v</I><TT><I><SUB>1</SUB></I></TT>< \
- > </I>+<I> m</I><TT><I><SUB>2</SUB></I></TT><I>v</I><TT><I><SUB>2<p>\
- </SUB></I></TT><p>\
- <TT><p>
-
- [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?<p>
- 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.<p>\
- Let the final velocity of the astronaut be <I>v</I>.<p>\
- momentum = mass x velocity<p>\
- initial momentum = final momentum<p>\
- 0 = 100<I>v</I> + 500 x 0.1<p>\
- Therefore,<p>\
- 100<I>v</I> = -50<p>\
- <img src="sa7q20a" align=center><p>\
- <center>= -0.5 m/s.</center><p>
-
-