A 30.0 kg hammer, moving with speed 20.0m/s, strikes a steel spike 2.30 cm in diameter….

Q: A 30.0 kg hammer, moving with speed 20.0m/s, strikes a steel spike 2.30 cm in diameter. The hammer rebounds with speed 10.0 m/s after 0.110s. What is the average stress in the spike during the impact?

(a) 1.97 × 107 N/m²

(b) 3.2 ×107 N/m²

(c) 4.6 × 107 N/m²

(d) 8.2 ×107 N/m²

Ans: (a)

Sol:
Force = Change in momentum per sec.

Change in momentum = 2 m u (since hammer rebounds with same speed)

$ \displaystyle F = \frac{2 m u}{t}$

$ \displaystyle Stress = \frac{F}{A} = \frac{F}{\pi r^2}$

Leave a Comment