Question

A bullet of mass 6.20 10-3 kg, moving at 1320 m/s impacts a tree stump and penetrates 11.00 cm into the wood before coming to rest. (Take the direction of the bullet to be in the positive direction.) (a) Assuming that the acceleration of the bullet is constant, find the force (including direction) exerted by the wood on the bullet.

Answer 1

Answer:

  F = -49.1   10³ N

Explanation:

Let's use the kinematics to find the acceleration the acceleration of the bullet that they tell us is constant

   $v_{f}$² = v₀² + 2 a x

Since the bullet is at rest, the final speed is zero

    x = 11.00 cm (1 m / 100 cm) = 0.110 m

    0 = v₀² + 2 a x

   a = -v₀² / 2 x

   a = -1320²/(2 0.110)

   a = -7.92 10⁶ m / s²

With Newton's second law we find the force

   F = m a

   F = 6.20 10⁻³ (-7.92 10⁶)

   F = -49.1   10³ N

The sign means that it is the force that the tree exerts to stop the   bullet