momentum

5. (a) Force on wall = -[0.5 x (-4 - 5)/0.4] N = 11.25 N
(b) Force on ground = -[0.5 x (0 - 5)/0.4] N = 6.25 N

6. (a) Use conservation of momentum
500 x 30 = 400 x 50 + 100v
where v is the ejected velocity of the engine
hence, v = -50 m/s
(b) Change of KE of the ejection
= (1/2).(100).(50^2 - 30^2) J = 80 000 J

7. Use momentum conservation
4 x 3 = 4v + 2v'
where v and v' are the speeds of the 4 kg ball and 2 kg ball respectively
i.e. 12 = 4v + 2v'
6 = 2v + v' -------------- (1)

Use energy conservation
(1/2).(4).(3^2) = (1/2).(4).v^2 + (1/2).(2).v'^2
i.e. 18 = 2v^2 + v'^2 ------------- (2)

solve equations (1) and (2) to find v and v'



2015-07-04 00:33:32 補充：
Q:may you write the formula for 5a and b?
A: Force = rate of change of momentum (Newton's 2nd Law)
Take the original direction of motion of the ball as +ve
Initial momentum = 0.5 x 5 kg.m.s
Final momentum = 0.5 x (-4) kg.m.s

2015-07-04 00:39:59 補充：
(cont'd)...
Hence, change of momentum = final momentum - initial momentum = [(0.5).(-4) - (0.5).(5)] kg.m/s = (0.5).(-4 -5) kg.m/s
Force on ball = [(0.5).(-4-5)]/0.4 N

2015-07-04 00:40:53 補充：
(cont'd)...
The question asks for the force on the wall. By Newton's 3rd Law (action and reaction law), the force on wall = force on ball, but in opposite direction
i.e. (force on wall) = -(force on ball)

2015-07-04 00:41:27 補充：
(cont'd)...
Therefore, force on wall = - [(0.5).(-4-5)]/0.4 N = 11.25 N
That is, the force on wall is in the SAME direction as the original motion direction of the ball.

2015-07-04 00:43:07 補充：
In 5b, the ball doesn't rebound (because of smooth ground), hence its final momentum is zero.

2015-07-04 00:45:15 補充：
Q:why not 6b be (1/2)(400)(50^2)+(.5)(100)(50^2)-(.5)(500)(30^2)?
A: My interpretation of "ejection" is the "ejected engine".
May be you are right if it means "the ejection process".

2015-07-04 00:51:59 補充：
Q:i find that v=1 or 3 and v' can be 4 or 0. which one should i choose ?
A: Clearly, you should use v = 1 m/s and v' = 4 m/s

2015-07-04 00:52:31 補充：
(cont'd)
The reasons are:
1.If use v = 3 m/s and v' = 0 m/s, these are the original velocities of the balls. It means there is NO collision at all, as there is NO transfer of momentum.This is clearly not the case.

2015-07-04 00:56:26 補充：
(cont'd)
2. After a collision, there must be a transfer of momentum from the 4 kg ball to the 2 kg ball. The final speed of the 4 kg ball CANNOT BE HIGHER than the 2 kg ball. That is , the condition v < v' needs to be satisfied.

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2015-07-03 20:51:13 補充：
why not 6b be (1/2)(400)(50^2)+(.5)(100)(50^2)-(.5)(500)(30^2)?

2015-07-03 20:58:57 補充：
sorry i find that v=1 or 3 and v' can be 4 or 0. which one should i choose ?