Momentum

2015-07-03 4:31 am
1.A ball of mass 2 kg travelling with a velocity of 4 m/s at 45 degree hits the smooth ground in 4s as shown below.

(a)What are the forces acting on the ball during collision?
(b)(i) Assume the collision is elastic. What is the speed of the ball after collision?

(ii) What should be the horizontal component of the final velocity of the ball? Why?

(iii) Calculate the force acting on the ground by the ball during collision.

2.A rubber is dropped vertically onto a 80g trolley originally moving with 3m/s. The final common speed is 1.6m/s. Given that it takes 1 s for the rubber to rest on the trolley.
(a) Find the mass of the rubber,
(b) Find the internal force acting on the rubber during collision and state the physical origin of the force.

3. Two balls of same mass of 2kg are travelling with velocities of 2m/s due east and 3m/s due north initially. At a certain instant, they collide with each other and stick together. The collision time is 0.5s.
(a)Find the total common velocity of the particle by drawing a momentum vector diagram.
(b)Find the change of KE of the system .

4. The velocity of a ball of mass 3kg after being hit by wall changes from 4m/s due east to 3m/s due north in 5s. Find the force exerted on the wall by the ball during collision.

回答 (1)

2015-07-03 7:40 am
✔ 最佳答案
1. Where is the diagram?

2. (a) Use momentum conservation
0.08 x 3 = (0.08 + m) x 1.6
where m is the mass of the rubber
solve for m gives m = 0.07 kg

(b) Force = 0.07 x 1.6/1 N = 0.112 N
The force is frictional force.

3. (a) The final momentum is the resultant of the individual momentum of the 2 balls. The momentum diagram is a right angle triangle. The hypotenus of which gives the final momentum.
Hence, final common velocity = 1.8 m/s

(b) Initial KE = [(1/2).(2).(2^2) + (1/2).(2).(3^2)] J = 13 J
Final KE = (1/2).(4).(1.8^2) J = 6.48 J
Hence, change of KE = (6.48 - 13) J = - -6.52 J

4. Change of velocity of ball = square-root[3^2 + 4^2] m/s = 5 m/s
Force = 3 x 5/5 N = 3 N


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