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January 13th, 2014, 05:59 PM  #1 
Newbie Joined: Jul 2008 Posts: 12 Thanks: 0  Mechanics, conservation of momentum
A block of mass 3kg travelling at 5m/s catches up another block of mass 7 kg travelling at 2 m/s along the same line and in the same direction. A) if after the collision the second block has increased its speed to 3 m/s, what is the speed of the first block? B) you are given that the difference between the speeds of the two blocks after the collision cannot be greater than the difference of the speeds after the collision must satisfy the inequalities 0.8<v<2.9 Part A is easy using the principle of conservation of momentum. What about part B? 
January 13th, 2014, 08:38 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,034 Thanks: 1393 
Can you correct the wording of part B? It doesn't make sense and it's unclear what it's asking.

January 14th, 2014, 02:06 AM  #3 
Newbie Joined: Jul 2008 Posts: 12 Thanks: 0  Re: Mechanics, conservation of momentum
Correction of part B A block of mass 3kg travelling at 5m/s catches up another block of mass 7 kg travelling at 2 m/s along the same line and in the same direction. A) if after the collision the second block has increased its speed to 3 m/s, what is the speed of the first block? B) you are given that the difference between the speeds of the two blocks after the collision cannot be greater than the difference of the speeds after the collision. Deduce that the speed v m/s of the first block after collision must satisfy the inequalities 0.8<v<2.9 Part A is easy using the principle of conservation of momentum. What about part B? 
January 14th, 2014, 04:12 AM  #4 
Math Team Joined: Sep 2007 Posts: 2,409 Thanks: 6  Re: Mechanics, conservation of momentum
You were not asked to repeat what you said but to clarify part B. In both your posts part B says "the difference in speeds after the collision cannot be greater than the difference in speeds after the collision. Was one of those supposed to be "before"?

January 14th, 2014, 04:56 AM  #5 
Newbie Joined: Jul 2008 Posts: 12 Thanks: 0  Re: Mechanics, conservation of momentum
Sorry for the confusion. The question is as follows. Correction of part B A block of mass 3kg travelling at 5m/s catches up another block of mass 7 kg travelling at 2 m/s along the same line and in the same direction. A) if after the collision the second block has increased its speed to 3 m/s, what is the speed of the first block? B) you are given that the difference between the speeds of the two blocks after the collision cannot be greater than the difference of the speeds before the collision. Deduce that the speed v m/s of the first block after collision must satisfy the inequalities 0.8<v<2.9 Part A is easy using the principle of conservation of momentum. What about part B? 
January 14th, 2014, 10:17 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 18,034 Thanks: 1393 
For part B, you should be asked to prove that 0.8 ? v ? 2.9. If the first block is halted by the collision, conservation of momentum implies that the second block's speed is increased to 29/7 m/s, which is impossible, as it would mean that the overall energy of the system has increased. Hence the first block's direction of movement is not changed by the collision. Obviously, the second block's direction of movement is not changed by the collision. Conservation of momentum implies the second block's speed after the collision is (29  3v)/7 m/s. Immediately after the collision, the first block's speed cannot exceed the second block's speed (assuming the first block can't then be passing through the second), so v ? (29  3v)/7, which implies v ? 2.9. Difference in speeds before collision is 3 m/s. Difference in speeds after collision is ((29  3v)/7  v) m/s. Hence the condition specified in part B implies (29  3v)/7  v ? 3, which implies 0.8 ? v. Combining these inequalities gives 0.8 ? v ? 2.9. 
January 14th, 2014, 02:10 PM  #7 
Newbie Joined: Jul 2008 Posts: 12 Thanks: 0  Re: Mechanics, conservation of momentum
Thanks skipjack


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