Collisions – problems and solutions
1. Object A (3 kg) moves at a speed of 8 m/s and object B (5 kg) moves at a speed of 4 m/s. If the collision between the object A and B is perfectly elastic, what is the velocity of object A and B after the collision?
Mass of object A (m1) = 3 kg
Mass of object B (m2) = 5 kg
The speed of object A (v1) = 8 ms–1
The speed of object B (v2) = 4 ms–1
Wanted: v1‘ and v2‘
If both objects have the different mass and the velocity of both objects after the collision is not known yet, then the velocity after collision calculated using this equation :
The direction of both objects is the same so that the velocity of both objects have the same sign. If both objects move in opposite direction then one velocity has plus sign and another velocity has minus sign.
The velocity of object A (vA) after the collision :
The velocity of object B (vB) after the collision :
Velocity of object A (v1‘) after collision is 3 m/s and velocity of object B (v2‘) after collision is 7 m/s.
2. A ball has a momentum of P, collide a wall and reflected. The collision is perfectly elastic and its direction perpendicular to the wall. What is the change of the ball’s momentum?
Mass of ball = m
The velocity of the ball before collision = v
Velocity of ball after collision = -v (ball reflected leftward)
Ball’s momentum before collision (po) = m v
Ball’s momentum after collision (pt) = m (-v) = – m v
Wanted: The change of the ball’s momentum
The change of momentum :
Δp = pt – po
Δp = – m v – m v
Δp = – 2 m v
Δp = -2p
Minus sign indicates the direction of the ball.
3. Two objects, A and B, have the same mass move in an opposite direction and collide with each other. A moves east a speed of V and B move west at a speed of 2V. If the collision is perfectly elastic, then what is the velocity of both objects after the collision.
Both objects have the same mass.
A moves east at speed of V
B moves west at speed of 2V
Wanted: The velocity of both objects after the collision
If both objects have the same mass and the collision is perfectly elastic, then after collision both objects change their velocity.
After the collision, A moves west at 2V and B moves east at V.
4. Two objects with the same mass move along a straight line in opposite direction as shown in the figure below. If the velocity of object 2 after the collision (v2‘) is 5 m/s rightward, then what is the magnitude of the velocity of the object 1 after the collision.
Mass of each object = m
Velocity of object 1 before collision (v1) = 8 m/s
Velocity of object 2 after collision (v2) = 10 m/s
Velocity of object 2 after collision (v2‘) = 5 m/s
Wanted : Velocity of object 1 after collision (v1‘)
The collision is partially elastic.
m1 v1+ m2 v2 = m1 v1’ + m2 v2’
m (v1 + v2) = m (v1’ + v2’)
v1 + v2 = v1’ + v2’
8 + 10 = v1’ + 5
18 = v1’ + 5
v1’ = 18-5
v1’ = 13 m/s
5. A 10-gram bullet fired at 100 m s-1 , collide a block of wood at rest. Mass of block is 490 gram. The collision is inelastic. What is the speed of the block and bullet after collision.
mass of bullet (m1) = 10 gram
Speed of bullet (v1) = 100 m/s
Mass of block (m2) = 490 gram
Speed of block (v2) = 0 m/s (block at rest)
Wanted : The speed of the block and bullet after collision
m1 v1 + m2 v2 = (m1 + m2) v’
(10)(100) + (490)(0) = (10 + 490) v’
1000 + 0 = 500 v’
1000 =500 v’
v’ = 1000 / 500
v’ = 2 m/s
6. A truck moves at 10 m/s collide a car moves at 20 m/s. After collision, both truck and car move together at the same speed. Mass of truck is 1400 kg and mass of car is 600 kg. What is the velocity of both truck and car after collision.
speed of truck (v1) = 10 m/s
speed of car (v2) = 20 m/s
mass of truck (m1) = 1400 kg
mass of car (m2) = 600 kg
Wanted : The velocity of both truck and car after collision
m1 v1 + m2 v2 = (m1 + m2) v
(1400)(10) + (600)(20) = (1400 + 600) v
14000 + 12000 = 2000 v
26000 = 2000 v
v = 13 m/s
7. A 20-gram bullet moves at 10 m/s in a horizontal direction, collide a 60-gram block rest on a floor. After collision, bullet and block move together with the same speed and the same direction. What is the speed of both bullet and block.
Mass of bullet (mP) = 20 gram = 0.02 kg
Mass of block (mB) = 60 gram = 0.06 kg
The initial speed of bullet (vP) = 10 m/s
The initial speed of block (vB) = 0
Wanted : The speed of bullet and block after collision (v’)
The collision is inelastic.
mP vP + mB vB = (mP + mB) v’
(0.02)(10) + (0.06)(0) = (0.02 + 0.06) v’
0.2 + 0 = 0.08 v’
0.2 = 0.08 v’
v’ = 0.2 / 0.08
v’ = 2.5 m/s
8. Two objects, A and B, have the same mass = 1.5 kg approach each other and collide. Speed of object A (vA) = 4 m/s and speed of object B (vB) = 5 m/s. If the collision is inelastic, what is the speed of both objects after collision.
Mass of object A (mA) = 1.5 kg
Mass of object B (mB) = 1.5 kg
Speed of object A before collision (vA) = 4 m/s (positive rightward)
Speed of object B before collision (vB) = -5 m/s (negative leftward)
Wanted : The speed of both objects after collision
mA vA + mB vB = (mA + mB) v’
(1.5)(4) + (1.5)(-5) = (1.5 + 1.5) v’
6 – 7.5 = (3) v’
-1.5 = (3) v’
v’ = -1.5 / 3
v’ = -0.5 m/s
Minus sign indicates that after collision, both objects move leftward, same direction as object B.
- What is a collision in the context of physics? Answer: In physics, a collision refers to an event where two or more objects come in close contact with each other, typically exerting a force upon each other, resulting in an exchange of energy and momentum.
- How are elastic and inelastic collisions different? Answer: In an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, momentum is conserved, but kinetic energy is not. Perfectly inelastic collisions are a subset where the colliding objects stick together after the collision.
- Why is momentum always conserved in collisions? Answer: Momentum is conserved in collisions due to the conservation of momentum principle, which states that the total momentum of a closed system remains constant unless acted upon by external forces.
- Is kinetic energy always conserved in collisions? Answer: No, kinetic energy is only conserved in elastic collisions. In inelastic collisions, some of the kinetic energy is transformed into other forms of energy, such as potential energy, sound, or heat.
- Why might two cars stick together after a collision? Answer: If two cars stick together after a collision, it’s an example of a perfectly inelastic collision. The cars remain together due to mechanical interlocking, deformation, or other forces that overcome their ability to rebound from each other.
- Can an object undergoing a collision experience a change in kinetic energy even if its speed remains unchanged? Answer: Yes, if an object changes its direction during a collision but not its speed, its velocity (a vector quantity) changes, but its speed (a scalar quantity) remains the same. Though the magnitude of its kinetic energy (dependent on speed) might not change, the object has undergone a change in energy distribution.
- Why do billiard balls bounce off each other when they collide? Answer: Billiard balls undergo nearly elastic collisions. When one ball strikes another, the kinetic energy and momentum are transferred, causing the balls to bounce apart. The design and material of billiard balls make them efficient at conserving kinetic energy in these interactions.
- If two objects of different masses collide and stick together, will their combined velocity be the average of their initial velocities? Answer: Not necessarily. The combined velocity will depend on the law of conservation of momentum. If one object is much more massive than the other, the combined velocity will be closer to the initial velocity of the more massive object.
- In the real world, are there truly “elastic” collisions? Answer: In practice, no collision is perfectly elastic. While some collisions, like those between billiard balls, come close, there’s always some energy lost to other forms, such as sound or heat.
How does an airbag reduce the impact of a collision in a car accident? Answer: Airbags increase the time taken for a person’s momentum to change upon collision. By doing so, the force experienced by the person (which is inversely proportional to the time over which momentum changes) is reduced, minimizing potential injury.