Ex7c

Q1  6 + 2 = 2p + q = 8 - 3 + 2 = 2p - 2q = -1

2p + q = 8

2p - 2q = -1

________________

3q = 9

q = 3

=> p = 2.5

The loss in kinetic energy is given by kinetic energy before - kinetic energy after (18 + 4) - (12.5 + 9 )

2

22 - 21.5

2

=0.25 J

Q3 Our table will be different this time, and we will have only 1 unknown in it, the problem is we don't know the value of e

(i) 10(3) + 1(5) = 0(3) + 5(q)

35 = 5 q

q = 7 0 - 7 = -7

10 - 1 9

(ii) q = 7 from before

(iii) for body 1 = 3(0) - 3(10) = -30Nsi

for body 2 = 5(7) - 5(1) = 30Nsi

As we would expect the impulses are equal but opposite

Q5 e = 1/2 2(10) + 3 (0) = 2p + 3q = 20 -1/2 = p - q

10

-10 = 2p - 2q

Simultaneous equations

2p + 3q = 20

2p - 2q = -10

-10 =

5q = 30

q = 6

p = 1

(ii) = 1/2 ((2(100) + 0) - (2(1) + 3(36))

= 1/2 (200 - 110)

= 1/2 (90)

= 45 J

(iii)

This time we must calculate a force therefore we need to find the acceleration, we have u (u = q = 6), we have v (v = 0) and t (t = 2) 0 = 6 + 2a

6 = -2a

a = -3m/s2

Q6

(a) a few methods here, find at the end of the fall using v2 = ..... then find the bounce velocity with -e = 0.8

(b) recoil is when the gun / cannon etc jumps back (as so momentum is conserved (see demo problems above))

Q7

find the v of B after the collision, use this as the u before the next one, again follow the algorithm given above,

Q8

Same as, this time we can get it in term of 1 unknown due to the ratio being declared, put the velocity after as 11q & 13q

Q9

conservation of momentum

mu - 3mv = 3q = u - 3v = 3q =>q = (u-3v)/3

coeff

-q = - e => e = q/ (u + v)

u - 3v

3 u + 3v

Q12

2 parts