Question

In a ballistics test, a 24 g bullet traveling horizontally at 1500 m/s goes through a 25-cm -thick 370 kg stationary target and emerges with a speed of 950 m/s . the target is free to slide on a smooth horizontal surface.

(a) How long is the bullet in the target?

Answer 1

The bullet is in the target for approximately 2.04 x 10⁻¹ seconds, which is calculated by dividing the thickness of the target by the average bullet speed (average of the initial and final speeds).

Step-by-step explanation:

To determine how long the bullet is in the target, we can use the equation which relates speed, distance, and time:

Speed = Distance / Time

From the problem, we are given:

• The initial speed of the bullet: 1500 m/s
• The final speed of the bullet after passing through the target: 950 m/s
• The thickness of the target: 25 cm (or 0.25 m)

Assuming the deceleration of the bullet is constant while it is inside the target, the average speed of the bullet can be calculated using the initial and final speeds:

Average Speed = (Initial Speed + Final Speed) / 2

Therefore:

Average Speed = (1500 m/s + 950 m/s) / 2

= 1225 m/s

Now we can solve for the time:

Time = Distance / Average Speed

= 0.25 m / 1225 m/s

Time = 2.04 x 10⁻⁴ s.

Thus, the bullet is in the target for approximately 2.04 x 10⁻⁴ seconds.