{"id":1786,"date":"2018-04-10T07:01:18","date_gmt":"2018-04-09T23:01:18","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1786"},"modified":"2023-08-16T22:08:35","modified_gmt":"2023-08-16T22:08:35","slug":"impulse-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/impulse-problems-and-solutions.htm","title":{"rendered":"Impulse \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

Impulse \u2013 problems and solutions<\/span><\/p>\n

1. A 0.5 kg ball free fall from a height of h1<\/sub> = 7.2 meters and reflected a height of h2<\/sub> = 3.2 meters. Acceleration due to gravity<\/a> = 10 m\/s2<\/sup>. Determine impulse<\/a>.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of ball (m) = 0.5 kg<\/span><\/p>\n

First height (h1<\/sub>) = 7.2 meter<\/span><\/p>\n

Second height (h2<\/sub>) = 3.2 meter<\/span><\/p>\n

Acceleration due to gravity (g) = 10 m\/s2<\/sup> <\/span><\/p>\n

Wanted :<\/u> Impulse (I)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Velocity of ball before collisions <\/u>(v<\/u>o<\/u><\/sub>)<\/u><\/span><\/p>\n

Velocity of ball before collision calculated using equation of free fall motion. Known :<\/u> Height (h) = 7.2 meters, acceleration due to gravity (g) = 10 m\/s2<\/sup>. Wanted :<\/u> Final velocity after collision.<\/span><\/p>\n

v2<\/sup> = 2 g h <\/span><\/p>\n

vo<\/sub>2<\/sup> = 2(10)(7.2) = 144<\/span><\/p>\n

vo<\/sub> = 2(10)(7.2) = 12 m\/s <\/span><\/p>\n

Velocity of ball before collision<\/a> (vo<\/sub>) = -12 m\/s. Minus sign indicates the direction of ball.<\/span><\/p>\n

Velocity of ball after collision <\/u>(v<\/u>t<\/u><\/sub>)<\/u><\/span><\/p>\n

Velocity of ball after collision calculated using equation of vertical motion. Known <\/u>: height (h) = 3.2 meters, acceleration due to gravity (g) = -10 m\/s2<\/sup>, final velocity at the maximum height (vt<\/sub>2<\/sup>) = 0. Wanted :<\/u> Initial velocity after collision between ball and floor.<\/span><\/p>\n

vt<\/sub>2<\/sup> = vo<\/sub>2<\/sup> + 2 g h<\/span><\/p>\n

0 = vo<\/sub>2<\/sup> + 2 (-10)(3.2)<\/span><\/p>\n

vo<\/sub>2<\/sup> = 64<\/span><\/p>\n

vo<\/sub> = \u221a64 = 8 m\/s <\/span><\/p>\n

Velocity of ball after collision (vt<\/sub>) is 8 m\/s<\/span><\/p>\n

Impulse (I)<\/u><\/span><\/p>\n

Impulse (I) = the change in momentum<\/a> (\u0394p)<\/span><\/p>\n

I = m (vt <\/sub>\u2013 vo<\/sub>) = (0.5)(8-(-12)) = (0.5)(8 + 12) = (0.5)(20) = 10 Newton second<\/span><\/p>\n

2. A 5-gram ball free fall from a height and strikes the floor. Acceleration due to gravity, g = 10 ms-2<\/sup>. Velocity of ball before collision is 6 ms-1<\/sup> and after collision, the ball is reflected upright at 4 m\/s. Determine the impulse.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass<\/a> of ball (m) = 5 gram = 0.005 kg<\/span><\/p>\n

Velocity of ball before collision (vo<\/sub>) = -6 m\/s<\/span><\/p>\n

Velocity of ball after collision (vt<\/sub>) = 4 m\/s<\/span><\/p>\n

Plus and minus sign indicates that the direction before collision is opposite with the direction after collision.<\/span><\/p>\n

Wanted:<\/u> Impulse (I)<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Impulse (I) = the total change in momentum (\u0394p).<\/span><\/p>\n

I = \u0394p = m vt<\/sub> \u2013 m vo <\/sub>= m (vt<\/sub> \u2013 vo<\/sub>) <\/span><\/p>\n

I = (0.005)(4 \u2013 (-6)) = (0.005)(4 + 6) = (0.005)(10) = 0.05 Newton second<\/span><\/p>\n

3. A 20-gram thrown with velocity of 4 m.s-1<\/sup> to the left. After colliding with the wall, the ball is reflected with velocity of v2 <\/sub>= 2 m.s-1<\/sup> to the right. Determine the impulse.<\/span><\/p>\n

Known :<\/u><\/span><\/p>\n

Mass of ball (m) = 20 gram = 0.020 kg\"Impulse<\/span><\/p>\n

Velocity of ball before collision (vo<\/sub>) = -4 m\/s (to the left)<\/span><\/p>\n

Velocity of ball after collision (vt<\/sub>) = +2 m\/s (to the right)<\/span><\/p>\n

Plus and minus sign indicates the opposite direction.<\/span><\/p>\n

Wanted :<\/u> Impulse<\/span><\/p>\n

Solution :<\/u><\/span><\/p>\n

Impulse (I) = the change in momentum (\u0394p) = m vt<\/sub> \u2013 m vo<\/sub> <\/span><\/p>\n

Impulse (I) = m (vt<\/sub> \u2013 vo<\/sub>) = 0.02 (2 \u2013 (-4)) <\/span><\/p>\n

Impulse (I) = 0.02 (2 + 4) = 0.02 (6) <\/span><\/p>\n

Impulse (I) = 0.12 Newton second.<\/span><\/p>\n

4. A 2 kg ball moving at 5 m\/s strikes a wall and bounces back with a velocity of -5 m\/s. Find the impulse exerted on the ball.<\/span>
\nSolution: Impulse = change in momentum = m(v\u2082 – v\u2081) = 2(-5 – 5) = -20 kg\u00b7m\/s<\/span><\/p>\n

5. A 3 kg object experiences a force of 6 N for 4 seconds. What is the impulse experienced by the object?<\/span>
\nSolution: Impulse = F\u0394t = 6 \u00d7 4 = 24 N\u00b7s<\/span><\/p>\n

6. A 0.5 kg object is at rest and then accelerated by an impulse of 10 N\u00b7s. What is the final velocity?<\/span>
\nSolution: Final velocity = Impulse \/ mass = 10 \/ 0.5 = 20 m\/s<\/span><\/p>\n

7. An impulse of 30 N\u00b7s acts on a 6 kg object. What is the change in velocity?<\/span>
\nSolution: \u0394v = Impulse \/ mass = 30 \/ 6 = 5 m\/s<\/span><\/p>\n

8. A 100 g ball moving at 3 m\/s is stopped by an impulse. What is the value of the impulse?<\/span>
\nSolution: Impulse = m(v\u2082 – v\u2081) = 0.1(0 – 3) = -0.3 kg\u00b7m\/s<\/span><\/p>\n

9. A 2 kg object experiences a 4 N force for 5 seconds. Find the impulse.<\/span>
\nSolution: Impulse = F\u0394t = 4 \u00d7 5 = 20 N\u00b7s<\/span><\/p>\n

10. An object of mass 3 kg is moving at 3 m\/s and comes to rest due to an impulse. Find the impulse.<\/span>
\nSolution: Impulse = m(v\u2082 – v\u2081) = 3(0 – 3) = -9 kg\u00b7m\/s<\/span><\/p>\n

11. A 5 kg object experiences a force of 2 N for 3 seconds. What is the impulse?<\/span>
\nSolution: Impulse = F\u0394t = 2 \u00d7 3 = 6 N\u00b7s<\/span><\/p>\n

12. An impulse of 8 N\u00b7s acts on a 4 kg object. What is the change in velocity?<\/span>
\nSolution: \u0394v = Impulse \/ mass = 8 \/ 4 = 2 m\/s<\/span><\/p>\n

13. A 1 kg ball moving at 2 m\/s strikes a wall and reverses its direction at the same speed. Find the impulse exerted on the ball.<\/span>
\nSolution: Impulse = m(v\u2082 – v\u2081) = 1(-2 – 2) = -4 kg\u00b7m\/s<\/span><\/p>\n

14. A 4 kg object experiences a 3 N force for 3 seconds. Find the impulse.<\/span>
\nSolution: Impulse = F\u0394t = 3 \u00d7 3 = 9 N\u00b7s<\/span><\/p>\n

15. An impulse of 10 N\u00b7s acts on a 2 kg object. What is the change in velocity?<\/span>
\nSolution: \u0394v = Impulse \/ mass = 10 \/ 2 = 5 m\/s<\/span><\/p>\n

16. A 3 kg object experiences a force of 5 N for 2 seconds. What is the impulse experienced by the object?<\/span>
\nSolution: Impulse = F\u0394t = 5 \u00d7 2 = 10 N\u00b7s<\/span><\/p>\n

17. An object of mass 4 kg is moving at 5 m\/s and comes to rest due to an impulse. Find the impulse.<\/span>
\nSolution: Impulse = m(v\u2082 – v\u2081) = 4(0 – 5) = -20 kg\u00b7m\/s<\/span><\/p>\n

18. A 2 kg object experiences a 6 N force for 3 seconds. Find the impulse.<\/span>
\nSolution: Impulse = F\u0394t = 6 \u00d7 3 = 18 N\u00b7s<\/span><\/p>\n

19. An impulse of 12 N\u00b7s acts on a 3 kg object. What is the change in velocity?<\/span>
\nSolution: \u0394v = Impulse \/ mass = 12 \/ 3 = 4 m\/s<\/span><\/p>\n

20. A 5 kg object experiences a force of 1 N for 4 seconds. What is the impulse?<\/span>
\nSolution: Impulse = F\u0394t = 1 \u00d7 4 = 4 N\u00b7s<\/span><\/p>\n

21. An impulse of 15 N\u00b7s acts on a 5 kg object. What is the change in velocity?<\/span>
\nSolution: \u0394v = Impulse \/ mass = 15 \/ 5 = 3 m\/s<\/span><\/p>\n

22. A 3 kg ball moving at 4 m\/s is stopped by an impulse. What is the value of the impulse?<\/span>
\nSolution: Impulse = m(v\u2082 – v\u2081) = 3(0 – 4) = -12 kg\u00b7m\/s<\/span><\/p>\n

23. A 2 kg object experiences a 5 N force for 1 second. Find the impulse.<\/span>
\nSolution: Impulse = F\u0394t = 5 \u00d7 1 = 5 N\u00b7s<\/span><\/p>\n

    \n
  1. What is impulse?<\/strong><\/span>\n