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7. A 1,2 kg hammer is used to hit a nail horizontally into a block of wood. The hammer is moving at 15 m.s¹ immediately before it collides with the nail and rebounds at 60% of that speed. The nail's resisting force is 9 000 N. Calculate how long the hammer is in contact with the nail.​

User Prabhu M
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Answer:

ETo calculate the duration of the hammer's collision with the nail, we need to find the time it takes for the hammer to come to a stop and then reverse direction.

The hammer's initial momentum before the collision is given by its mass and velocity:

p_i = m * v_i = 1.2 kg * 15 m/s = 18 kg m/s

The hammer's final velocity after the collision can be found using the law of conservation of momentum, which states that the total momentum of an isolated system remains constant. We can write this as:

p_f = m * v_f = m * (-v_i * 0.6) = m * (-15 m/s * 0.6) = 9 kg m/s

The force experienced by the nail during the collision can be calculated as:

F = ma = (m * (v_f - v_i)) / t = (m * (9 kg m/s - 18 kg m/s)) / t

Since we know the force experienced by the nail is 9 000 N, we can use this to solve for t:

9 000 N = (1.2 kg * (9 kg m/s - 18 kg m/s)) / t

t = (1.2 kg * (9 kg m/s - 18 kg m/s)) / 9000 N

t = 0.00133 seconds

So the hammer is in contact with the nail for approximately 0.00133 seconds.

variable:

m: mass of the hammer, 1.2 kg

v_i: initial velocity of the hammer before the collision, 15 m/s

v_f: final velocity of the hammer after the collision, -9 m/s (negative sign indicates the direction of motion is opposite to the initial velocity)

t: duration of the hammer's collision with the nail, approximately 0

User Varun Malhotra
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