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Problem 60:  Newton’s cradle ( ) 1999 Paper II

N particles P1, P2, P3, , PN with masses m, qm, q2m, , qN1m, respectively, are at rest at distinct points along a straight line in gravity-free space. The particle P1 is set in motion towards P2 with velocity V and in every subsequent impact the coefficient of restitution is e, where 0 < e < 1.

Show that after the first impact the velocities of P1 and P2 are 1 eq 1 + q V    and     1 + e 1 + qV,


Show that, if q e, then there are exactly N 1 impacts.
Show further that, if q = e, then the total loss of kinetic energy after all impacts have occurred is 1 2me1 eN1V 2.


This situation models the toy called ‘Newton’s Cradle’ which consists of four or more heavy metal balls suspended from a frame so that they can swing. At rest, they are in contact in a line. When the first ball is raised and let swing, there follows a rather pleasing pattern of impacts. In this case, the coefficient of restitution is nearly 1 and the balls all have the same mass, so, as the first displayed formula shows, the impacting ball is reduced to rest by the impact. At the first swing of the ball, nothing happens except that the first ball is reduced to rest and the last ball swings away. Note that this is consistent with the balls being separated by a very small amount; what actually happens is that the ball undergoes a small elastic deformation at the impact, and the impulse takes a small amount of time to be transmitted across the ball to the next ball.

Solution to problem 60

(i) Let V 1 and V 2 be the velocities of P1 and P2 after the first collision. Using conservation of momentum and Newton’s law of impact at the first collision results in the equations

mV = mV 1 + mqV 2,V 1 V 2 = eV ,


V 1 = 1 eq 1 + q V    and    V 2 = 1 + e 1 + qV .

(ii) Note that V 1 is positive since eq e2 < 1. Conserving momentum and using Newton’s law (and doing a bit of algebra) shows that the speeds V ̂2 and V 3 of P2 and P3 after the next collision are given by

V ̂2 = 1 eq 1 + q V 2 = 1 eq 1 + q 1 + e 1 + qV    and    V 3 = 1 + e 1 + qV 2 = 1 + e 1 + q2V. ()

Note that V ̂2 V 1 so there is no further collision between P1 and P2. Applying this argument at each collision shows that there are exactly N 1 collisions: P1 with P2;  P2 with P3;  etc.

(iii) The speed of Pk after it has hit Pk+1 is (by extending ())

1 eq 1 + q 1 + e 1 + qk1V,

and the speed of Pk+1 after this collision and before it hits Pk+2 is

1 + e 1 + qkV.

If q = e,

V 1 = 1 e2 1 + e V = (1 e)V

and similarly the final speeds of P2, , PN1 are all (1 e)V . The final speed of PN is V . Thus the final total kinetic energy is

1 2(m + mq + + mqN2)[(1 e)V ]2 + 1 2mqN1V 2 = 1 2m1 eN1 1 e (1 e)2V 2 + 1 2meN1V 2 = 1 2m(1 e + eN)V 2,

(replacing all the q’s with e’s and summing the geometric progression). Thus the loss of kinetic energy is

1 2mV 2 1 2m(1 e + eN)V 2,

as required.


There are two tricky aspects to these multiple collision questions. First there is the matter of notation. Above, I have used a hat for the second collision of a particle (V ̂2), retaining the subscript for labelling particles. That works, but if P2 undergoes another collision, entailing a double hat, it starts getting messy. You could use a different letter, but that gets confusing. The only good method is to use a double subscript: V m,n is the velocity of the mth particle after the nth collision. But that is a sledge hammer for this nut of a question.

The other tricky aspect is getting the signs right. I always think of velocity (not speed), and it is always positive for particles travelling to the right. I also use common sense to check each equation! As usual, a good diagram of each collision is essential.