A student investigates the effect of a crumple zone on the force exerted during a collision The student has one trolley with a s
pring at the front and another trolley without a spring
silver foil
spring
silver foil
trolley
The student uses the arrangement in Figure 10.
trolley
silver foil
ramp
trolley
silver foil
rigid wall
too
Figure 10
000 ms
millisecond timer
After a trolley is released, it accelerates down a slope and bounces off a rigid wall
The speed of a trolley can be measured just before a collision with the wall and just after a collision with the wall.
The silver foils are connected to a millisecond timer.
The silver foils make contact with each other during the collision, so the time they are in contact can be read from the millisecond timer.
Explain how the student could investigate the effect of a crumple zone on the average force exerted during the collision.
Your explanation should include:
how to determine the force (you may wish to refer to an equation) how the effect of crumple zones may be shown in the investigation precautions that may be necessary to achieve accurate results.
(6)
silver foil
spring
silver foil
trolley
The student uses the arrangement in Figure 10.
trolley
silver foil
ramp
trolley
silver foil
rigid wall
too
Figure 10
000 ms
millisecond timer
After a trolley is released, it accelerates down a slope and bounces off a rigid wall
The speed of a trolley can be measured just before a collision with the wall and just after a collision with the wall.
The silver foils are connected to a millisecond timer.
The silver foils make contact with each other during the collision, so the time they are in contact can be read from the millisecond timer.
Explain how the student could investigate the effect of a crumple zone on the average force exerted during the collision.
Your explanation should include:
how to determine the force (you may wish to refer to an equation) how the effect of crumple zones may be shown in the investigation precautions that may be necessary to achieve accurate results.
(6)
1 answer:
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Answer:
kinetic Energy = 12.58 GJ
Explanation:
1 metric ton is equal to 1000 kg
then,
93000 metric ton
Mass is m = 93000 x 1000 kg
speed is v = 32 knots
1 knot is 0.514 m/s
then ,
v = 32 x 0.514 = 16.448 m/s
To solve for the Kinetic Energy (KE), we have;
KE = 0.5 x m x v²
KE = 0.5*93000*1000*(16.448)²
= 12.58 x J
= 12.58 GJ