Question

A 7.91g bullet is fired into a 1.52-kg ballistic pendulum initially at rest and becomes embedded in it. The pendulum subsequently rises a vertical distance of 6.89 cm. What was the initial speed of the bullet?
a. 0.38 km/s
b. 0.44 km/s
c. 0.50 km/s
d. 0.54 km/s
e. 0.024 km/s

Answer 1

Answer:

D

Explanation:

By law of conservation of momentum:

momentum before collision = momentum after collision

$m_{1} u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\\ m_1 = mass of bullet\\u_1 = initial speed of bullet\\m_2 = mass of pendulum\\u_2 = initial speed of pendulum\\v_1 = final speed of bullet\\v_2 = final speed of pendulum$

Initial speed of bullet is unknown whereas initial speed of pendulum will be zero as it was at rest.

Final speed of bullet and pendulum will be equal as bullet is embedded in pendulum and both moves together a vertical distance of 6.89cm.

Using third equation of motion:

$2 a S = v_{f}^2 - v_{i}^2$

where:

$a = -g = acceleration due to gravity = 9.8 m/s^2 ( for vertical motion)\\S = h = 6.89cm = Vertical height or distance it covers\\v_f = final speed = 0 (after covering 6.89cm)\\v_i = initial speed (unknown)$

Thus by placing values $v_i = 1.1620 m/s$

this speed will be final speed of collision for the calculation of initial speed of bullet.

Putting values:

$(7.91) * x + (1.52*10^3) * 0 = (7.91 + 1.52*10^3) * 1.162\\7.91 x = 1775.431\\x = 224m/s$

This 224m/s = 0.224Km/s which is closest to D