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}
ΔEp = mgΔh
ΔEp = 0.602 × 9.8 × 0.086
ΔEp = 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, Ek = ½ mv2
Ek = ½ × 0.002 ×3902
Ek = 152 J or 153 J or 150 J
(ii) The kinetic energy is not the same. So, the collision is inelastic.
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