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:\[\begin{align*} I &= p_2 – p_1\\ I& = m (v_2 – v_1)\\ I& = 0,1 (20 – 0) = 2 Ns\end{align*}\]
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:
\[\begin{align*} I &= p_2 – p_1\\ I& = m (v_2 – v_1)\\ I &= 0,05 (-4 – 6)\\ I &= 0,05 (-10) = -0,5 Ns\end{align*}\](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:
\[\small \begin{align*} (m_A+m_B+m_p)v &= m_Bv'_B+(m_A+m_p)v'_B\\ (50+30+50)(10) &= 30(5)+(50+50)v'_A\\ 1300&= 150+(100)v'_A\\ 100v'_A&=1150\\v'_A&=\frac{1150}{100}=11,5ms^{-1} \end{align*}\]
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.
\[\begin{align*} I &=\left (\frac{1}{2} \right )t\times (AD+BC)\\ I &=\left (\frac{1}{2} \right )(2)\times (4+2)\\ I &=6Ns \end{align*}\]
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...
\[\begin{align*} I &= m (v_2-v_1)\\ F\Delta t &=m (v_2-v_1)\\F(0,1) &=0,5(100-0)\\ F &=500N \end{align*}\]
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 $v'_A = v'_B = v'$\[\begin{align*} m_Av_A+m_Bv_B &= m_Av'_A+m_Bv'_B\\ m_Av_A+m_Bv_B &= (m_A+m_B) v'\\ 0.6(2)+0.4(0)&= (0.6+0.4)v'\\ 1,2&=v'\\v'&=1,2ms^{- 1} \end{align*}\]
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:\[\small \begin{align*} m_1v_1+m_2v_2 &= m_1v'_1+m_2v'_2\\ 0.25(20)+0.25(-10)&= 0.25v'_1+ 0.25v'_2\\ 5-2.5&=0.25v'_1+0.25v'_2\\v'_1+v'_2&=10 ....(1) \end{align*}\] Restitution coefficient:\[\begin{align*} e &= -\frac{(v'_1-v'_2)}{(v_1-v_2)}\\ 1 &= -\frac{(v'_1-v '_2)}{20-(-10)}\\v'_1-v'_2&=-30 ....(2) \end{align*}\]eliminate both equations:\[\begin{align*} v'_1+v'_2&=10.......(1)\\v'_1-v'_2&=-30....(2)+\\\hline 2v'_1&=-20\\v'_1&=-10ms^{-1}\end{align*}\]to equation (1):\[\begin{align*} v'_1+v'_2&=10\\-10+v'_2&=10\\v'_2&=20ms^{-1}\end{align*}\]WOW way from Mr Dimpun: \[\begin{align*} c&=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\c&=\frac{0,25(20)+0,25(-10)}{0,25+0 ,25}\\c&=\frac{5-2.5}{0,5}=5\\ \end{align*}\]\[\begin{align*} v'_1&=c+e(c -v_1)\\v'_1&=5+1(5-20)\\ v'_1&=-10ms^{-1}\\ \\ v'_2&=c+e(c-v_2)\\v' _2&=5+1(5-(-10))\\ v'_2&=20ms^{-1} \end{align*}\]
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:\[\small \begin{align*} v' &= \sqrt{2gh}\\v'&=\sqrt{2(10)(0,2)}=2ms^{-1} \end{align*}\]Law of Conservation of Momentum:\[\small \begin{align*} v' &= \sqrt{2gh}\\v'&=\sqrt{2(10)(0,2)}=2ms^{-1} \end{align*}\]Law of Conservation of Momentum:\[\small \begin{align*} m_pv_p+m_bv_b &= (m_p+m_b)v'\\ (0,01)v_p+(2,49)0 &= (0,01+2,49)v'\\0,01v_p &= 2,5(2)\\0,01v_p &= 5\\ v_p&=500ms^{-1} \end{align*}\]
QUESTIONS AND SOLUTIONS IMPULTS AND MOMENTUM
When B jumps back:
\[\small \begin{align*} (m_A+m_B+m_p)v &= m_Bv'_B+(m_A+m_p)v'_B\\ (50+30+50)(10) &= 30(5)+(50+50)v'_A\\ 1300&= 150+(100)v'_A\\ 100v'_A&=1150\\v'_A&=\frac{1150}{100}=11,5ms^{-1} \end{align*}\]
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.
\[\begin{align*} I &=\left (\frac{1}{2} \right )t\times (AD+BC)\\ I &=\left (\frac{1}{2} \right )(2)\times (4+2)\\ I &=6Ns \end{align*}\]
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...
\[\begin{align*} I &= m (v_2-v_1)\\ F\Delta t &=m (v_2-v_1)\\F(0,1) &=0,5(100-0)\\ F &=500N \end{align*}\]
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 $v'_A = v'_B = v'$\[\begin{align*} m_Av_A+m_Bv_B &= m_Av'_A+m_Bv'_B\\ m_Av_A+m_Bv_B &= (m_A+m_B) v'\\ 0.6(2)+0.4(0)&= (0.6+0.4)v'\\ 1,2&=v'\\v'&=1,2ms^{- 1} \end{align*}\]
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:\[\small \begin{align*} m_1v_1+m_2v_2 &= m_1v'_1+m_2v'_2\\ 0.25(20)+0.25(-10)&= 0.25v'_1+ 0.25v'_2\\ 5-2.5&=0.25v'_1+0.25v'_2\\v'_1+v'_2&=10 ....(1) \end{align*}\] Restitution coefficient:\[\begin{align*} e &= -\frac{(v'_1-v'_2)}{(v_1-v_2)}\\ 1 &= -\frac{(v'_1-v '_2)}{20-(-10)}\\v'_1-v'_2&=-30 ....(2) \end{align*}\]eliminate both equations:\[\begin{align*} v'_1+v'_2&=10.......(1)\\v'_1-v'_2&=-30....(2)+\\\hline 2v'_1&=-20\\v'_1&=-10ms^{-1}\end{align*}\]to equation (1):\[\begin{align*} v'_1+v'_2&=10\\-10+v'_2&=10\\v'_2&=20ms^{-1}\end{align*}\]WOW way from Mr Dimpun: \[\begin{align*} c&=\frac{m_1v_1+m_2v_2}{m_1+m_2}\\c&=\frac{0,25(20)+0,25(-10)}{0,25+0 ,25}\\c&=\frac{5-2.5}{0,5}=5\\ \end{align*}\]\[\begin{align*} v'_1&=c+e(c -v_1)\\v'_1&=5+1(5-20)\\ v'_1&=-10ms^{-1}\\ \\ v'_2&=c+e(c-v_2)\\v' _2&=5+1(5-(-10))\\ v'_2&=20ms^{-1} \end{align*}\]
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:\[\small \begin{align*} v' &= \sqrt{2gh}\\v'&=\sqrt{2(10)(0,2)}=2ms^{-1} \end{align*}\]Law of Conservation of Momentum:\[\small \begin{align*} v' &= \sqrt{2gh}\\v'&=\sqrt{2(10)(0,2)}=2ms^{-1} \end{align*}\]Law of Conservation of Momentum:\[\small \begin{align*} m_pv_p+m_bv_b &= (m_p+m_b)v'\\ (0,01)v_p+(2,49)0 &= (0,01+2,49)v'\\0,01v_p &= 2,5(2)\\0,01v_p &= 5\\ v_p&=500ms^{-1} \end{align*}\]
