A bullet of mass 2.0 g is fired horizontally into a block of wood of mass 600 g.

 Question 43

A bullet of mass 2.0 g is fired horizontally into a block of wood of mass 600 g. The block is suspended from strings so that it is free to move in a vertical plane.

 

The bullet buries itself in the block. The block and bullet rise together through a vertical distance of 8.6 cm, as shown in Fig. 3.1.

 

Fig. 3.1

 

(a) (i) Calculate the change in gravitational potential energy of the block and bullet. [2]

 

(ii) Show that the initial speed of the block and the bullet, after they began to move off together, was 1.3 m s^-1. [1]

 

 

(b) Using the information in (a)(ii) and the principle of conservation of momentum, determine the speed of the bullet before the impact with the block. [2]

 

 

(c) (i) Calculate the kinetic energy of the bullet just before impact. [2]

 

(ii) State and explain what can be deduced from your answers to (c)(i) and (a)(i) about the type of collision between the bullet and the block. [2]

 

 

 

 

 

Reference: Past Exam Paper – June 2015 Paper 2 Q3

 

 

 

 

 

Solution:

(a) (i)

{Total mass = mass of bullet + mass of block = 2 + 600 = 602 g = 0.602 kg}

ΔE_p = mgΔh

ΔE_p = 0.602 × 9.8 × 0.086

ΔE_p = 0.51 J

 

 

(ii)

{KE of bullet and block = Change in gravitational PE

½ mv² = 0.51 J}

v² = (2 × 0.51) / 0.602

Initial speed of block and bullet, v = 1.3 m s^-1 

 

 

(b)

{Sum of momentum before impact = Sum of momentum after impact

Before the impact, the bullet is moving while the block is at rest.

Momentum of bullet before impact = mv = 2.0 × v

Sum of momentum before impact = 2v + 0 = 2v

After the impact, the bullet and the block move as a single body with the same speed.

Momentum after impact = (600+2) × 1.3

Sum of momentum before impact = Sum of momentum after impact }

2v = 602 × 1.3             (allow 600)

Speed of bullet, v = 390 m s^-1

 

 

(c) (i)

Kinetic energy of bullet, E_k = ½ mv²

E_k = ½ × 0.002 ×390²

E_k = 152 J        or 153 J           or 150 J

 

 

(ii) The kinetic energy is not the same. So, the collision is inelastic.