1. Two train cars are on the same track. Car A is 2800 kg ,and is traveling west
ID: 1434393 • Letter: 1
Question
1. Two train cars are on the same track. Car A is 2800 kg ,and is traveling west at 20 m/s, while car B is stopped on the tracks. Car B has a mass of 1500 kg. After the collision both cars are stuck together, what is their combined velocity?
2. Two children are riding in bumper cars. Car A has a mass of 455 kg, and is traveling 2.0 m/s to the right, and car B is 440 kg and is traveling at 3.5 m/s to the left. When they collide car A is traveling left at 1 m/s, what is the velocity of car B after the collision?
3. Two children are on an ice rink. Child A, who has a mass of 55 kg, is traveling right at 2.0 m/s when she collides with child B, who has a mass of 60 kg, and is traveling at 1.5 m/s to the left. When they collide they stay together. What is their final velocity?
4. A 2 kg blob of clay moving at 8 m/s to the left impacts a 4 kg blob of clay moving at 6 m/s to the right. They stick together after the collision. What is the final velocity of the larger blob of clay?
5. Two carts with masses of 4.5 kg and 4.0 kg move toward each other on a frictionless track with speeds of 5.0 m/s and 4.0 m/s. The carts stick together after the collision. Find the final speed of both carts.
6. In 1987, Marisa Canofoglia, of Italy, roller-skated at a record setting speed of 11.9 m/s. If the magnitude of her momentum was 6.60*10^2 kg*m/s, what was her mass?
7. The first human satellite,Sputnik, had a mass of 83.6 kg and a momentum of 6.63*10^5 kg*m/s. What was the satllite's speed?
8. The specially designed armored car that was built for Leonid Brezhnev when he was head of the Soviet Union had a mass of about 6000 kg. Suppose this car is accelerated from rest by a force of 8000 N to the east. What is the car's velocity after 8.0 sec?
Explanation / Answer
1). Mass of car A, mA= 2800kg
Velocity of car A, uA= 20m/s (Assuming West direction to be positive)
Mass of car B, mB= 1500kg
Velocity of car B, uB= 0m/s (Since it was at rest)
After collision they both stick together. Let there final velocity be "v", then following Conservation of Momentum Principle,
mAuA+mBuB= (mA+mB)v
using given values in above,
(2800)(20)+(1500)(0)= (2800+1500)v
v= 56000/4300= 13.023 m/s (ANS)
2). Mass of car A, mA= 455kg
Initial Velocity of car A, uA= 2.0m/s (Assuming right direction to be positive)
Mass of car B, mB= 440kg
Initial Velocity of car B, uB= - 3.5m/s (Then left direction will be negative)
Final Velocity of car A, vA= -1.0m/s (Left direction will be negative)
Let there final velocity of car B be "vB", then following Conservation of Momentum Principle,
mAuA+mBuB= mAvA+mBvB
using given values in above,
(455)(2.0)+(440)(-3.5)= (455)(-1.0)+(440)vB
vB= (910-1540+455)/(440)= -0.3977 m/s (ANS)
3). Mass of child A, mA= 55kg
Initial Velocity of child A, uA= 2.0m/s (Assuming right direction to be positive)
Mass of child B, mB= 60kg
Initial Velocity of child B, uB= - 1.5m/s (Then left direction will be negative)
After collision they both stick together. Let there final velocity be "v", then following Conservation of Momentum Principle,
mAuA+mBuB= (mA+mB)v
using given values in above,
(55)(2.0)+(60)(-1.5)= (55+60)v
v= (110-90)/(115)= 0.174 m/s (ANS)
4). Mass of 1st clay bob, m1= 2kg
Initial Velocity of 1st clay bob, u1= -8.0m/s (Assuming right direction to be positive, then left direction will be negative)
Mass of 2nd clay bob, m2= 4kg
Initial Velocity of 2nd clay bob, u2= 6m/s
After collision they both stick together. Let there final velocity be "v", then following Conservation of Momentum Principle,
m1u1+m2u2= (m1+m2)v
using given values in above,
(2)(-8.0)+(4)(6.0)= (2+4)v
v= (-16+24)/(6)= 1.33 m/s (ANS)
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