Question

A rifle, which has a mass of 5.50 kg., is used to fire a bullet, which has a massof m = 65.0 grams., at a "ballistics pendulum". The ballistics pendulum consistsof a block of wood, which has a mass of M = 5.00 kg., attached to two stringswhich are L = 125 cm long. When the block is struck by the bullet the blockswings backward until the angle between the ballistics pendulum and thevertical reaches a maximum angle ofa= 38.0o.a. What will be the maximum gravitational energy contained in the ballisticspendulum when it reaches the maximum angle?b. What was the velocity of the block of wood immediately after being struck bythe bullet?c. What was the velocity of the bullet immediately before it strikes the block ofwood?d. How much work was done by the bullet as it lodged in the block of wood?e. What will be the recoil velocity of the rifle?f. How much energy was released when the bullet was fired?

Answer 1

Answer:

Part a)

$U = 13 J$

Part b)

$v = 2.28 m/s$

Part c)

$v = 177.66 m/s$

Part d)

$W = 1012.7 J$

Part e)

$v = 2.1 m/s$

Part f)

$E = 1037.2 J$

Explanation:

Part a)

As we know that the maximum angle deflected by the pendulum is

$\theta = 38^o$

so the maximum height reached by the pendulum is given as

$h = L(1 - cos\theta)$

so we will have

$h = L(1 - cos38)$

$h = 1.25(1 - cos38)$

$h = 0.265 m$

now gravitational potential energy of the pendulum is given as

$U = mgh$

$U = 5(9.81)(0.265)$

$U = 13 J$

Part b)

As we know that there is no energy loss while moving upwards after being stuck

so here we can use mechanical energy conservation law

so we have

$mgh = \frac{1}{2}mv^2$

$v = \sqrt{2gh}$

$v = \sqrt{2(9.81)(0.265)}$

$v = 2.28 m/s$

Part c)

now by momentum conservation we can say

$mv = (M + m) v_f$

$0.065 v = (5 + 0.065)2.28$

$v = 177.66 m/s$

Part d)

Work done by the bullet is equal to the change in kinetic energy of the system

so we have

$W = \frac{1}{2}mv^2 - \frac{1}{2}(m + M)v_f^2$

$W = \frac{1}{2}(0.065)(177.66)^2 - \frac{1}{2}(5 + 0.065)2.28^2$

$W = 1012.7 J$

Part e)

recoil speed of the gun can be calculated by momentum conservation

so we will have

$0 = mv_1 + Mv_2$

$0 = 0.065(177.6) + 5.50 v$

$v = 2.1 m/s$

Part f)

Total energy released in the process of shooting of gun

$E = \frac{1}{2}Mv^2 + \frac{1}{2}mv_1^2$

$E = \frac{1}{2}(5.50)(2.1^2) + \frac{1}{2}(0.065)(177.6^2)$

$E = 1037.2 J$