If m is the mass of the object and v is its velocity (also a vector quantity), then the momentum of the object is: p=m.v.
In SI units, momentum is measured in meters kilograms per second (kg m/s).
We have learned about momentum, impulses and collisions in the article labeled momentum . Now, let's deepen your understanding with a variety of other questions.
Problems And Solution Of Momentum And Impulses
Problems no. 1. A ball of mass 0.1 kg is initially at rest, then after being hit with a stick and the speed of the ball becomes 20 m/s. Calculate the magnitude of the impulse of the hitting force!
Momentum Solution:I=p2–p1I=m(v2–v1)I=0,1(20–0)=2Ns
Problems no. 2 A ball with a mass of 50 grams is thrown horizontally with a speed of 6 m/s to the right, the ball hits a wall and is reflected with a speed of 4 m/s to the left. Calculate the magnitude of the impulse exerted by the wall on the ball!
Momentum Solution:
I=p2–p1I=m(v2–v1)I=0,05(−4–6)I=0,05(−10)=−0,5Ns(a negative sign indicates that the ball reverses direction, moves to the left)
Problems no. 3 Two people are in a boat with a mass of 50 kg which is moving to the right with a speed of 10 m/s. If A has a mass of 50 kg and B has a mass of 30 kg, then calculate the speed of the boat when B jumps backwards with a speed of 5 m/s!
Momentum Solution:
QUESTIONS AND SOLUTIONS IMPULTS AND MOMENTUM
When B jumps back:
(mA+mB+mp)v=mBv′B+(mA+mp)v′B(50+30+50)(10)=30(5)+(50+50)v′A1300=150+(100)v′A100v′A=1150v′A=1150100=11,5ms−1
Problems no. 4 Look at the picture below!
The figure above is a curve of the force versus time, which acts on a particle of mass 2 kg initially. The impulse of the force is...
Momentum Solution:
Impulse is equal to the area under the F-vs-t graph. In the figure, the impulse of the force is the area of the trapezoid ABCD.
I=(12)t×(AD+BC)I=(12)(2)×(4+2)I=6Ns
Problems no. 5 A ball of mass 0.5 kg is at rest and is hit until the ball slides with a speed of 100 m/s. If the length of the batsman touching the ball is 0.1 seconds, then the batsman's great force is...
I=m(v2−v1)FΔt=m(v2−v1)F(0,1)=0,5(100−0)F=500N
Problems no. 6 A ball A moving to the right with a speed of 2 m/s strikes a stationary ball B, if after the collision balls A and B come together, then calculate the velocity of each ball after the collision!
Momentum Solution:
After colliding the two balls merge then v′A=v′B=v′mAvA+mBvB=mAv′A+mBv′BmAvA+mBvB=(mA+mB)v′0.6(2)+0.4(0)=(0.6+0.4)v′1,2=v′v′=1,2ms−1
Problems no. 7 Two balls are moving towards each other, where the first ball moves to the right with a speed of 20 m/s and the second ball moves to the left with a speed of 10 m/s so that a perfectly elastic collision occurs. If each ball has a mass of 0.25 kg, then calculate the speed of the two balls after they collide!
Momentum Solution:
Law of conservation of momentum:m1v1+m2v2=m1v′1+m2v′20.25(20)+0.25(−10)=0.25v′1+0.25v′25−2.5=0.25v′1+0.25v′2v′1+v′2=10....(1) Restitution coefficient:e=−(v′1−v′2)(v1−v2)1=−(v′1−v′2)20−(−10)v′1−v′2=−30....(2)eliminate both equations:v′1+v′2=10.......(1)v′1−v′2=−30....(2)+2v′1=−20v′1=−10ms−1to equation (1):v′1+v′2=10−10+v′2=10v′2=20ms−1WOW way from Mr Dimpun: c=m1v1+m2v2m1+m2c=0,25(20)+0,25(−10)0,25+0,25c=5−2.50,5=5v′1=c+e(c−v1)v′1=5+1(5−20)v′1=−10ms−1v′2=c+e(c−v2)v′2=5+1(5−(−10))v′2=20ms−1
Problems no. 8 A bullet of mass 0.01 kg is fired at a block of mass 2.49 kg suspended by a string as shown in the figure below.
If after the collision the bullet is embedded in the block, and the position of the block increases by h = 20 cm, then calculate the speed of the bullet when it hits the block!
Momentum Solution:
The object and the bullet are raised to a height of h = 0.2m, then:v′=√2ghv′=√2(10)(0,2)=2ms−1Law of Conservation of Momentum:v′=√2ghv′=√2(10)(0,2)=2ms−1Law of Conservation of Momentum:mpvp+mbvb=(mp+mb)v′(0,01)vp+(2,49)0=(0,01+2,49)v′0,01vp=2,5(2)0,01vp=5vp=500ms−1
QUESTIONS AND SOLUTIONS IMPULTS AND MOMENTUM
When B jumps back:
(mA+mB+mp)v=mBv′B+(mA+mp)v′B(50+30+50)(10)=30(5)+(50+50)v′A1300=150+(100)v′A100v′A=1150v′A=1150100=11,5ms−1
Problems no. 4 Look at the picture below!
The figure above is a curve of the force versus time, which acts on a particle of mass 2 kg initially. The impulse of the force is...
Momentum Solution:
Impulse is equal to the area under the F-vs-t graph. In the figure, the impulse of the force is the area of the trapezoid ABCD.
I=(12)t×(AD+BC)I=(12)(2)×(4+2)I=6Ns
Problems no. 5 A ball of mass 0.5 kg is at rest and is hit until the ball slides with a speed of 100 m/s. If the length of the batsman touching the ball is 0.1 seconds, then the batsman's great force is...
I=m(v2−v1)FΔt=m(v2−v1)F(0,1)=0,5(100−0)F=500N
Problems no. 6 A ball A moving to the right with a speed of 2 m/s strikes a stationary ball B, if after the collision balls A and B come together, then calculate the velocity of each ball after the collision!
Momentum Solution:
After colliding the two balls merge then v′A=v′B=v′mAvA+mBvB=mAv′A+mBv′BmAvA+mBvB=(mA+mB)v′0.6(2)+0.4(0)=(0.6+0.4)v′1,2=v′v′=1,2ms−1
Problems no. 7 Two balls are moving towards each other, where the first ball moves to the right with a speed of 20 m/s and the second ball moves to the left with a speed of 10 m/s so that a perfectly elastic collision occurs. If each ball has a mass of 0.25 kg, then calculate the speed of the two balls after they collide!
Momentum Solution:
Law of conservation of momentum:m1v1+m2v2=m1v′1+m2v′20.25(20)+0.25(−10)=0.25v′1+0.25v′25−2.5=0.25v′1+0.25v′2v′1+v′2=10....(1) Restitution coefficient:e=−(v′1−v′2)(v1−v2)1=−(v′1−v′2)20−(−10)v′1−v′2=−30....(2)eliminate both equations:v′1+v′2=10.......(1)v′1−v′2=−30....(2)+2v′1=−20v′1=−10ms−1to equation (1):v′1+v′2=10−10+v′2=10v′2=20ms−1WOW way from Mr Dimpun: c=m1v1+m2v2m1+m2c=0,25(20)+0,25(−10)0,25+0,25c=5−2.50,5=5v′1=c+e(c−v1)v′1=5+1(5−20)v′1=−10ms−1v′2=c+e(c−v2)v′2=5+1(5−(−10))v′2=20ms−1
Problems no. 8 A bullet of mass 0.01 kg is fired at a block of mass 2.49 kg suspended by a string as shown in the figure below.
If after the collision the bullet is embedded in the block, and the position of the block increases by h = 20 cm, then calculate the speed of the bullet when it hits the block!
Momentum Solution:
The object and the bullet are raised to a height of h = 0.2m, then:v′=√2ghv′=√2(10)(0,2)=2ms−1Law of Conservation of Momentum:v′=√2ghv′=√2(10)(0,2)=2ms−1Law of Conservation of Momentum:mpvp+mbvb=(mp+mb)v′(0,01)vp+(2,49)0=(0,01+2,49)v′0,01vp=2,5(2)0,01vp=5vp=500ms−1
