2.3 Solutions
81
From (5)
m 1
m 2
= α
2
=
√
2 − 1
2 = 3 − 2
√
2
2.32 (i) If the common velocity of the merged bodies is v then momentum conservation gives
(m A + m B )v = m A v A + m B v B
∴ v =
m A v A + m B v B
m A + m B
(ii) v =
3
2
m B (5 ˆ
i + 3 ˆ
j) + m B (− ˆ
i + 4 ˆ
j)
3
2
m B + m B
= 2.6 ˆ
i + 3.4 ˆ
j
(iii) p A = m A (υ − υ A ) = m A
2.6 ˆ
i + 3.4 ˆ
j − (5 ˆ
i + 3 ˆ
j)
= m A
−2.4 ˆ
i + 0.4 ˆ
j
p A = 1200
(−2.4) 2 + (0.4) 2 = 2920 N m
F A =
p A
t
=
2920
0.2
= 14,600 N
p B = m B (
υ −
υ B ) = m B
2.6 ˆ
i + 3.4 ˆ
j − (− ˆ
i + 4 ˆ
j)
= m B
3.6 ˆ
i − 0.6 ˆ
j
m B =
2
3
m A =
2
3
× 1200 = 800 kg
p B = 800
(3.6) 2 + (−0.6) 2 = 2920 N m
F B =
p B
t
=
2920
0.2
= 14,600 N
(iv) K =
1
2
(m A + m B )υ 2 =
1
2
(1200 + 800)
(2.6)
2
+ (3.4)
2
= 18, 320 J
2.33 Let the velocity of the particle moving below the x-axis be v’. Momentum
conservation along x- and y-axis gives
mv 0 = mv cos θ + mv
cos β
(1)
0 = mv sin θ − mv
sin β
(2)