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A2 physics - basic MOMENTUM question?

14) The student carries out an experiment to investigate momentum and collisions. A ball bearing is rolled down a ramp clamped at the edge of a bench. The ball bearing makes a head-on collision with a marble at the bottom of the ramp.

The ball bearing and the marble both move forward in the initial direction of the ball bearing after the collision, landing in a tray containing a layer of sand.

(a) Initially the ball bearing rolls down the ramp without the marble in a place. The ball bearing falls a verticle distance of 1.2m and lands a horizontal distance 95cm forward of the end of the ramp. Show that the speed of the ball bearing leaving the ramp is about 2ms^-1 (worked it out to be 1.9ms^-1 which is correct, also obtained that t = 0.5)

(b) When the ball bearing is rolled down the ramp with the marble in place, it knocks the marble forwards. The ball bearing lands a horizontal distance 64cm forward of the end of the ramp, and the marble lands a further 93cm forward from this positio. Calculate the ratio of masses of the ball bearing and he marble (ok all I know is that conservation of momentum is involved. but I just don't know where to begin!)

any help is HIGHLY appreciated. i know it's a chunky question but i really need to know the concepts behind this kinda stuff :redface:

thanks in advance :biggrin:
Reply 1
We know the ball bearing's (m1) initial velocity (u1) as 1.9m/s
We need its final velocity (v1) so calculate final vertical velocity probably by using v2=u2+2as v^2 = u^2 + 2as and its horizontal velocity (distance/time). You now need to find it's net velocity so add the vectors together.
Do the same for the marble (m2) i.e calculate final vertical velocity, horizontal velocity and add them together (v1)

Plug values into:
m1u1+m2u2=m1v1+m2v2 m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2
But since m2 is stationary to begin with u2 is 0 m2u2 is 0.

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