Can someone answer these couple of questions. Ive answered them and my questions different ot the BOB. Checked over them and they appear right (probs wrong).
1) A child rolls a 50 gram marble up a playground slide that is inclined at 15 degrees to the horizontal. The slide is 3.5m long and the marble is launched with a speed of 4.8m/s.
How fast is the marble travelling when it is halfway up the slide?
2) A car travelling with a constant speed of 80km/hr passes a stationary motorcycle policeman. The cop sets off in pursuit, acclerating uniformly to 80km/hr in 10secs, and 100km/hr after a further 5 secs. he stays at 100km/hr for the rest of the journey. At what time will the policeman catch up with the car?
I know the distance of both need to be the same and you solve for time, but i keep getting a different answer.
3) Cassie starts from rest at the top of a 3.2m long smooth playground slide and slides to thwe bottom with a constant acceleration. If she takes 2.4 secs to reach the bottom. Calculate her average acceleration?
Last bit. If your asked to find the speed of a golf ball which has fallen 2 meters and find the speed when it has rebounded. How is this done. I got the right answer but im trying to think why?
For speed would it be just pick a point that is jsut above the point of impact (ie. 0.5 meters) assuming that down is engative? So youve picked a point on its upwards journey.
And for velocity, the displacement would be say 1.5m from the top?
Im just kind of making a educated guess here ^.
1. u=4.8, a=10sin(15)=-2.6, s=1.75, v=?
v^2=u^2+2as=23.04-9.06=13.98
v=3.74 ms^-1
You can also use energy conservation like this:
Bottom: GPE=0, KE=1/2 mv^2=0.576
1/2 way up slide: GPE=mgh=mg(1.75sin(15))=0.23
KE=0.576-0.23=0.35=1/2 mv^2
v^2=14
v=3.74 ms^-1
2. 80 km/hr=22.2 ms^-1
100 km/hr=27.8 ms^-1
In the first 15 seconds:
Car travels 22.2*15=333.3 m
Policeman travels 1/2*10*22.2 + 1/2*5*(22.2+27.
=236.1 m
After exactly 15 seconds the policeman is behind by 97.2 m, and the difference in velocity is 27.8-22.2=5.6 ms^-1
The policeman takes a further 97.2/5.6=17.5 s to catch up, so total time taken is 15+17.5=32.5 s
3. u=0, t=2.4, s=3.2, a=?
s=ut+1/2 at^2
3.2=2.88a
a=1.11 ms^-2
For the last question, you can find out the KE just before the ball hits the ground (=GPE at top=mgh=20m). Assuming an elastic collision (I think that's what you're supposed to assume), the KE just after the ball bounces is also 20m. This means that in both cases, the speed can be calculated via:
20m=KE=1/2 mv^2
v^2=40
v=6.32 ms^-1
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