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Urgent help m1!!!!!!!!

Really struggling on this question.
Two particles, A and B, of mass 0.2kg and 0.5kg respectively, are connected by a light inextensible string passing over a fixed smooth light pulley. The particles are released from rest with the string taut and the hanging parts vertical. Then, at the instant when A and B are moving with speed 6.3ms^-1, B strikes an inelastic horizontal floor from which it does not rebound.

Calculate the total time from the instant when the particles were first released until the instant when A first comes to instantaneous rest. (You may assume that A does not reach the pulley.) the correct answer I must get is 2.41s.

I've calculated the magnitude of the impulse exerted on the floor by B, which is 3.15 N s.

I have calculated that a=4.2, tension=2.8.

(edited 10 years ago)

Reply 1

ghostwalker

Original post by joel abraham1111

...

There are two parts to the motion, where the acceleration is constant within each part.

Before B hits the ground. So you need to work out how long it takes from rest for A to reach 6.3 m/s

After B has splodged on the ground, A is now moving freely under gravity with an initial upwards vertical speed of 6.3 m/s (for this section of the working). So how long does it take to come to rest.

And add the two.

Reply 2

joel abraham1111

i have got 1.5m/s for one of the parts however i have no idea how to calculate the second.

Reply 3

ghostwalker

Original post by joel abraham1111

i have got 1.5m/s for one of the parts however i have no idea how to calculate the second.

I presume you mean 1.5 s.

For the second part, the initial velocity is 6.3m/s, final velocity 0 m/s and acceleration is "-g". That's sufficient info. to calculate how long it takes to come to instantaneous rest.

The correct answer is 2.14, not 2.41 (as in your first post)