Question

Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward.(See the figure below) The measured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a 4.20 g bullet intoa 2.50 kg block hanging by a thin wire 75.0 cm long, causing the block to swing upward to a maximum angle of 31.2 degrees from the vertical.

What was the original speed of this bullet?
THANK YOU

Answer 1

mass of the bullet m = 4.2 gmass of the block is M = 2.5 kgLet v be the speed of the bulletThe initial momentum of the bullet P_b = mvThe initial momentum of the block is P_B = 0The final momentum is P' = (m+M)v'Where v' is the velocity of the bullet and block after the bullet is embeddedBy applying the conservation of momentum,mv = (m+M)v'The bullet mass system has kinetic energy at the bottom K = 0.5(m+M)v'²The bullet mass system haspotential energy at thetop U = (m+M)ghWhere h is the vertical displacement of the block mass systemh = L - Lcosθwhere L is the length of the string = 75 cmθ is the angle made with the vertical = 31.2^oU = (m+M)gL(1-cosθ)Applying the conservation of energy0.5(m+M)v'² = (m+M)gL(1-cosθ)v' = √[2gL(1-cosθ)]v' = 1.458 m/sThereforemv = (m+M)v'(0.0042 kg)v = (0.0042 kg + 2.5 kg)(1.458 m/s)v = 869.315 m/sThe initial speed of the bullet is 869.315 m/s

Answer 2

According to conservation of energy,
1/2mV^2 = (m+M)gh
here h = l(1-casθ) = 0.75*0.144 = 0.1084 m
1/2*0.0042 V^2 = 2.5042*9.8*0.1084
V^2 = 1266.791
V = 35.59 m/s
the original speed of the bullet is 35.59 m/s.