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Impulse
A stone S is sliding on ice. The stone is moving along a straight line ABC, where AB = 24 m and AC = 30 m. The stone is subject to a constant resistance to motion of magnitude 0.3 N. At A the speed of S is 20 m s–1, and at B the speed of S is 16 m s–1. Calculate
(a) the deceleration of S,
(b) the speed of S at C.
(c) Show that the mass of S is 0.1 kg.
At C, the stone S hits a vertical wall, rebounds from the wall and then slides back along the line CA. The magnitude of the impulse of the wall on S is 2.4 N s and the stone continues to move against a constant resistance of 0.3 N.
(d) Calculate the time between the instant that S rebounds from the wall and the instant that S comes to rest.
a.) a=-3m/s², Deceleration of S is 3m/s²
b.) v=14.83m/s
c.) Proven m=0.1kg
d.) I have worked it out both ways, however - Do you have to take <-- as the positive direction because the Impulse is 2.4? Or can you work it the other way round by taking I=-2.4 and --> as the positive direction?
(a) the deceleration of S,
(b) the speed of S at C.
(c) Show that the mass of S is 0.1 kg.
At C, the stone S hits a vertical wall, rebounds from the wall and then slides back along the line CA. The magnitude of the impulse of the wall on S is 2.4 N s and the stone continues to move against a constant resistance of 0.3 N.
(d) Calculate the time between the instant that S rebounds from the wall and the instant that S comes to rest.
a.) a=-3m/s², Deceleration of S is 3m/s²
b.) v=14.83m/s
c.) Proven m=0.1kg
d.) I have worked it out both ways, however - Do you have to take <-- as the positive direction because the Impulse is 2.4? Or can you work it the other way round by taking I=-2.4 and --> as the positive direction?
Newton
Since the impulse is received in the direction which is opposite to the initial direction of motion, -2.4=mv-mu.
Newton.
Newton.
Ah, okay, thanks Newton.
Queen_A
I'm really confused as to how to answer part c). After you find out v usinng 2/4=mv-mu, you then have u=38.8322....,v=0, a? . How do you find the acceleration? (well, usinf F=ma, but how do you know P? (If P-0.3=ma)...thanks).
Oh..and what's the answer
Oh..and what's the answer
a) (v^2)=(u^2)+2as
=>a=(((16^2)-(20^2))/48)
=>a=-3 m(s^(-1))
b) (v^2)=(u^2)+2as
=>v=sqrt[(16^2)-36]
=>v=14.8 m(s^(-1))
c) The only force acting on the particle is friction in the reverse direction of its motion
F=ma
=>-0.3=-3m
=>m=0.1 kg
d) I=m(v-u)
=>24=v-14.8
=>v=-9.2 m(s^(-1))
v=u+at=>0=9.2-3t
=>t=3.1 s
Newton.
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