Answer: 1a. To find the recoil velocity on the staple gun after firing the staple, we need to use the principle of conservation of momentum. This principle states that the total momentum of an isolated system remains constant if no external forces act upon it.
The momentum of an object is given by the formula p = mv, where m is the mass and v is the velocity of the object.
Let's assume the mass of the staple is m_s and the velocity of the staple after being fired is v_s. The mass of the staple gun is m_g and the velocity of the staple gun after firing the staple is v_g.
According to the principle of conservation of momentum, the initial momentum of the system (before firing the staple) is zero, and the final momentum of the system (after firing the staple) is:
m_s * v_s + m_g * v_g = 0
We know the mass of the staple gun is 0.910 kg, the mass of staple is 6.8g, velocity of the staple is 1.23 m/s. So we can calculate the recoil velocity on the staple gun after firing the staple by solving the above equation:
(m_s * v_s) + (m_g * v_g) = 0
(6.8 * 10^-3 kg * 1.23 m/s) + (0.910 kg * v_g) = 0
v_g = -(6.8 * 10^-3 kg * 1.23 m/s) / 0.910 kg = -0.0075 m/s
The recoil velocity on the staple gun after firing the staple is -0.0075 m/s (negative sign indicates that the recoil velocity is in the opposite direction of the staple)
1b. The recoil velocity is so small because the mass of the staple is much smaller than the mass of the staple gun. Therefore, the staple gun will only experience a small recoil velocity due to the firing of the staple. This is because the principle of conservation of momentum states that the momentum of an object is directly proportional to its mass and velocity. In this case, the velocity of the staple is much larger than the velocity of the staple gun, but the mass of the staple is much smaller than the mass of the staple gun, which results in a small recoil velocity.