{"id":1799,"date":"2018-04-11T08:44:24","date_gmt":"2018-04-11T00:44:24","guid":{"rendered":"https:\/\/gurumuda.net\/physics\/?p=1799"},"modified":"2023-08-09T08:44:49","modified_gmt":"2023-08-09T08:44:49","slug":"momentum-problems-and-solutions","status":"publish","type":"post","link":"https:\/\/gurumuda.net\/physics\/momentum-problems-and-solutions.htm","title":{"rendered":"Momentum \u2013 problems and solutions","gt_translate_keys":[{"key":"rendered","format":"text"}]},"content":{"rendered":"

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

1. 0.5-kg ball moves at 2 m\/s. Then the ball is hit with force <\/span>F <\/span>opposite to the ball <\/span>direction<\/span>, so the ball speed is changed to 6 m\/s<\/span>. <\/span>The ball in contact with a hitter for <\/span>0.01 second, what is the change in momentum<\/a> of the ball. <\/span><\/p>\n

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

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

Initial velocity (vo<\/sub>) = 2 m\/s<\/span><\/p>\n

Final velocity (vt<\/sub>) = -6 m\/s<\/span><\/p>\n

Time interval (t) = 0.01 second<\/span><\/p>\n

Wanted :<\/u> The change in momentum<\/span><\/p>\n

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

\u2206p = m vt<\/sub> \u2013 m vo<\/sub> = m (vt <\/sub>\u2013 vo<\/sub>) <\/span><\/p>\n

\u2206p = (0.5 kg)(- 6 m\/s \u2013 2 m\/s) <\/span><\/p>\n

\u2206p = (0.5 kg)(-8) <\/span><\/p>\n

\u2206p = 4 kg m\/s<\/span><\/p>\n

2. A 100-gram object moves at 5 m\/s. Force F works for 0.2 seconds to stop the object. Determine magnitude of force F.<\/span><\/p>\n

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

Mass<\/a> of object (m) = 100 gram = 100\/1000 = 0.1 kg<\/span><\/p>\n

Initial velocity (vo<\/sub>) = 5 m\/s<\/span><\/p>\n

Final velocity (vt<\/sub>) = 0<\/span><\/p>\n

Time interval (\u0394t) = 0.2 seconds<\/span><\/p>\n

Wanted :<\/u> Magnitude of force F<\/span><\/p>\n

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

I = \u0394P<\/span><\/p>\n

F (\u0394t) = m (vt<\/sub> \u2013 vo<\/sub>)<\/span><\/p>\n

F (0.2) = 0.1 (0 \u2013 5)<\/span><\/p>\n

F (0.2) = 0.1 (\u2013 5)<\/span><\/p>\n

F (0.2) = -0,5<\/span><\/p>\n

F = -0,5 \/ 0.2<\/span><\/p>\n

F = -2.5 N<\/span><\/p>\n

Magnitude of force F = 2.5 Newton. Minus sign indicates the direction of force opposite with the direction of the object. <\/span><\/p>\n

3. A 100-gram ball free fall from the height of 20 cm without initial velocity and then the ball hits the floor. After collision, the ball is reflected upward (acceleration due to gravity is 10 ms-2<\/sup>). What is the change in momentum of ball.<\/span><\/p>\n

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

Mass of ball (m) = 100 gram = 0.1 kg<\/span><\/p>\n

Height (h) = 20 cm = 0.2 meters<\/span><\/p>\n

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

Velocity of ball after hits the floor (vt<\/sub>) = 1 m\/s <\/span><\/p>\n

Wanted :<\/u> The change in momentum<\/span><\/p>\n

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

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

Calculate velocity of ball before collision using equation of free fall motion. Known :<\/u> height of ball (h) = 0.2 meters, acceleration due to gravity (g) = 10 m\/s2<\/sup>. Wanted :<\/u> velocity of ball when hits the floor. <\/span><\/p>\n

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

v2<\/sup> = 2 (10)(0.2) = 4 <\/span><\/p>\n

v = \u221a4 = -2 m\/s<\/span><\/p>\n

Minus sign indicates that the direction of ball before collision is opposite with the direction of ball after collision.<\/span><\/p>\n

The change in momentum of ball (\u0394p) <\/u><\/span><\/p>\n

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

\u0394p = (0.1)(1 \u2013 (-2)) = (0.1)(1 + 2) = (0.1)(3) = 0.3 Newton second<\/span><\/p>\n

4. An object initially at rest, explosion into 2 parts with a ratio of 3 : <\/span>2<\/span>. The larger part of the mass is thrown at a speed of 20 m\/s. <\/span>What is t<\/span>he velocity of the smaller <\/span>part.<\/span><\/span><\/p>\n

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

Mass of object 1 before explosion = m<\/span><\/p>\n

Velocity of object 1 before explosion = 0 (object at rest)<\/span><\/p>\n

Mass of the larger part after explosion (m1<\/sub>) = 3m<\/span><\/p>\n

Mass of the smaller part after explosion (m2<\/sub>) = 2m<\/span><\/p>\n

Velocity of the larger part after explosion (v1<\/sub>‘) = 20 m\/s<\/span><\/p>\n

Wanted :<\/u> Velocity of the smaller part after explosion (v2<\/sub>‘)<\/span><\/p>\n

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

The equation of the law of the conservation of momentum :<\/span><\/p>\n

m1 <\/sub>v1 <\/sub>= m1 <\/sub>v1<\/sub>‘ + m2 <\/sub>v2<\/sub>‘<\/span><\/p>\n

(m)(0) = (3m)(20) + (2m) v2<\/sub>‘<\/span><\/p>\n

0 = 60m + (2m) v2<\/sub>‘<\/span><\/p>\n

60m = -2m v2<\/sub>‘<\/span><\/p>\n

60 = -2 v2<\/sub>‘<\/span><\/p>\n

v2<\/sub>‘ = -60\/2<\/span><\/p>\n

v2<\/sub>‘ = -30 m\/s<\/span><\/p>\n

Minus sign indicates that the direction of the smaller part is opposite with the direction of the larger part.<\/span><\/p>\n

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