Solutions

1

a) p = mv = .003 kg x 1500 m/s = 4.5 N s. Since the ball must have 4.5 N s momentum, the 0.145 kg ball must be moving 31.0 m/s.
b) Comparing kinetic energies the bullet has 3.38 kJ while the ball has 69.7 J or 0.0697 kJ. The bullet has more kinetic energy.

2

Since the forces applied are all internal, the total change in momentum is zero. Before the collision, the momentum is (0.2 kg)(55 m/s) = 11.0 N s. After the collision the club has
(0.2 kg)(40 m/s) = 8.0 N s of momentum. The ball must have the other 3.00 N s of momentum. Therefor (0.046 kg)v = 3.00 N s. Solving we find the speed to be 65.2 m/s.

3

In an elastic collision both momentum and energy are conserved. The total momentum before the collision is
(10 g)(20 cm/s) + (15 g)(-30 cm/s) = -250 d s.
The energy before the collision is
0.5(10 g)(20 cm/s)² + 0.5(15 g)(30 cm/s)² = 8.75 x 10³ ergs.
-250 = (10 g)v₁ + (15 g)v₂
and
8750 = 0.5(10 g)v₁² + 0.5(15 g)v₂².
Solving we find
v₁ = -40 cm/s and v₂ = +10 cm/s.