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