Question

Impact Time and Impact Force Bouncy Board Purpose To investigate the effect of stopping time when momentum changes Apparatus table meterstick short length (30 to 40 cm) of weak string (mailing parcel twine or equivalent) various masses Discussion In bungee jumping itis important that the cord supporting the jumper stretches.ifthe cord has no stretch, then when fully extended it either brings the jumper to a sudden halt or the cord snaps. In either case, ouch! Whenever a falling object is brought to a halt, the force that slows the fall depends on the time it acts. You will see evidence of this in this activity. Procedure Through the small hole at the end of a meterstick, thread a piece of string about 30 cm long and tie it in place. (If there is no hole, consider drilling one, or tying the string very tightly) Attach a mass to the other end of the string--try a kilogram for Figure 1 starters. Place the stick on a tabletop and slide the stick so that most of its length extends over the edge of the table, Figure 1. While holding the stick firmly to the table, hold the mass slightly beyond the edge of the stick, then drop the mass. The string and the bending of the stick should stop its fall. The string shouldn't break (if it does break, try a stronger string or a smaller mass-experiment). Now repeat this activity, but with only a small portion of tfie stick extending over the table's edge. See Figure 2. What happens now? Figure 2 Try different configurations to see what condikions result in string breaking. Experiment with different masses, different amounts of meterstick overhang, and different string lengths 35 BOUNCY BOARD

Answer 1

By heating gradually it's rate of change of momentum decreases hence force experience by bungee Jumper reduces As we know Ft = m delta v implies F = m delta v/t by increasing t we can reduce force Because of high impulse string breaks longer string can strech more and hence reduces maximum force and this maximum force is limited by breaking strength of string since we know that F = m delta v/t implies F proportionalto m hence by doubling mass force will double To reduce maximum force