1. A 2300 kg truck moving at 22 m/s collides with a stationary 1400 kg passenger car, and the two stick together. Assuming that they are free to move afterward, what is the momentum of the truck+car object after the collision?
a) 14 kg m/s
b) 31000 kg m/s
c) 51000 kg m/s
d) 19000 kg m/s
e) 81000 kg m/s
2. A 5.0 kg ball of clay traveling at 10 m/s collides with a 10 kg ball of clay traveling in the same direction at 2.0 m/s. What is their combined speed if the two balls stick together when they touch?
a) 4.7 m/s
b) 1.3 m/s
c) 1.0 m/s
d) 70 m/s
e) 15 m/s
3. A steel ball traveling at 6 m/s hits a stationary line of three steel balls. Which statement is true if the collision is perfectly elastic?
a) The original ball bounces back at -6 m/s while the other three balls remain stationary.
b) The original ball stops and a single ball moves to the right at 6 m/s.
c) The original ball bounces back at -3 m/s and one ball moves to the right at +3 m/s.
d) All four balls move to the right at 1.5 m/s
e) The original ball bounces back at -6 m/s and one ball moves to the right at +6 m/s.
4. One sphere of mass 1 kg is moving at 5 m/s to the right until it collides with a stationary, 2 kg sphere. After the collision, both spheres travel to the right: the 1 kg sphere at 1 m/s, and the 2 kg sphere at 2 m/s. What kind of collision took place?
a) an elastic collision
b) a perfectly inelastic collision
c) an inelastic (but not perfectly inelastic) collision
d) This situation is impossible.
e) More information is needed.
5. Ball A has a mass of 4.0 kg. Ball B has twice the mass of ball A. The two balls collide and stick together in a perfectly inelastic collision. After the collision the combined balls are at rest. If the velocity of ball A before the collision was 6.7 m/s, what was the velocity of ball B before the collision?
a) -6.7 m/s
b) 13 m/s
c) -3.4 m/s
d) 3.4 m/s
e) 0.0 m/s
1. The momentum of an object is given by the equation p = mv, where p is the momentum, m is the mass, and v is the velocity.
Before the collision, the momentum of the truck is given by p1 = (mass of truck)(velocity of truck) = (2300 kg)(22 m/s) = 50600 kg m/s.
Before the collision, the momentum of the car is given by p2 = (mass of car)(velocity of car) = (1400 kg)(0 m/s) = 0 kg m/s.
After the collision, assuming they stick together, the momentum of the truck+car object is given by p_final = (mass of truck + mass of car)(velocity of truck+car).
Since they stick together, the final velocity of the truck+car object is the same as the initial velocity of the truck, which is 22 m/s.
Therefore, p_final = (2300 kg + 1400 kg)(22 m/s) = 3700 kg * 22 m/s = 81400 kg m/s.
The momentum of the truck+car object after the collision is 81400 kg m/s.
Therefore, the correct answer is option e) 81000 kg m/s.
2. Again, before the collision, the momentum of the first ball is given by p1 = (mass of first ball)(velocity of first ball) = (5 kg)(10 m/s) = 50 kg m/s.
Before the collision, the momentum of the second ball is given by p2 = (mass of second ball)(velocity of second ball) = (10 kg)(2 m/s) = 20 kg m/s.
After the collision, assuming they stick together, the total momentum of the combined balls is given by p_final = (mass of first ball + mass of second ball)(velocity of combined balls).
Since they stick together, the final velocity of the combined balls is the same for both balls.
Therefore, p_final = (5 kg + 10 kg)(velocity of combined balls).
To find the combined velocity, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Therefore, 50 kg m/s + 20 kg m/s = (15 kg)(velocity of combined balls).
Simplifying the equation, we get 70 kg m/s = 15 kg * velocity of combined balls.
Dividing both sides of the equation by 15 kg, we get velocity of combined balls = 70 kg m/s / 15 kg = 4.7 m/s.
Therefore, the combined speed of the balls after the collision is 4.7 m/s.
The correct answer is option a) 4.7 m/s.
3. In a perfectly elastic collision, both kinetic energy and momentum are conserved.
Since the original ball is traveling with a velocity of 6 m/s, its momentum is given by p1 = (mass of original ball)(velocity of original ball) = (1 steel ball)(6 m/s) = 6 kg m/s.
After the collision, the original ball bounces back, so its velocity is reversed and becomes -6 m/s. Therefore, its momentum is -6 kg m/s.
The total momentum after the collision is equal to the total momentum before the collision, so the momentum of the three stationary balls combined is 6 kg m/s.
Since the three balls are stationary, their individual momenta are 0 kg m/s.
Therefore, the correct answer is option a) The original ball bounces back at -6 m/s while the other three balls remain stationary.
4. In an elastic collision, both momentum and kinetic energy are conserved.
Before the collision, the momentum of the 1 kg sphere is given by p1 = (mass of 1 kg sphere)(velocity of 1 kg sphere) = (1 kg)(5 m/s) = 5 kg m/s.
Before the collision, the momentum of the 2 kg sphere is given by p2 = (mass of 2 kg sphere)(velocity of 2 kg sphere) = (2 kg)(0 m/s) = 0 kg m/s.
After the collision, the momentum of the 1 kg sphere is given by p1_final = (mass of 1 kg sphere)(final velocity of 1 kg sphere) = (1 kg)(1 m/s) = 1 kg m/s.
After the collision, the momentum of the 2 kg sphere is given by p2_final = (mass of 2 kg sphere)(final velocity of 2 kg sphere) = (2 kg)(2 m/s) = 4 kg m/s.
Since momentum is conserved, the sum of the final momenta must be equal to the sum of the initial momenta.
Therefore, 1 kg m/s + 4 kg m/s = 5 kg m/s = (1 kg + 2 kg)(final velocity).
Simplifying the equation, we get final velocity = 5 kg m/s / 3 kg = 1.67 m/s.
Since the final velocities of both spheres are to the right, the collision is not perfectly inelastic.
Therefore, the correct answer is option c) an inelastic (but not perfectly inelastic) collision.
5. In a perfectly inelastic collision, both momentum and kinetic energy are conserved.
Before the collision, the momentum of ball A is given by p_A = (mass of ball A)(velocity of ball A) = (4 kg)(6.7 m/s) = 26.8 kg m/s.
Since the combined balls are at rest after the collision, their velocity is 0 m/s.
After the collision, the momentum of the combined balls is given by p_final = (mass of ball A + mass of ball B)(velocity of combined balls).
Since the mass of ball B is twice the mass of ball A, the mass of the combined balls is (4 kg + 2*4 kg) kg = 12 kg.
Therefore, p_final = (12 kg)(0 m/s) = 0 kg m/s.
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision.
Therefore, 26.8 kg m/s = 0 kg m/s + (12 kg)(velocity of combined balls).
Simplifying the equation, we get 26.8 kg m/s = 12 kg * velocity of combined balls.
Dividing both sides of the equation by 12 kg, we get velocity of combined balls = 26.8 kg m/s / 12 kg = 2.23 m/s.
Since the velocity of ball B is in the same direction as the velocity of ball A, the velocity of ball B before the collision is also 2.23 m/s.
Therefore, the correct answer is option d) 3.4 m/s.