an object of mass 2 kg travelling along a straight line with a velocity of 5 m/s collides with the end is fixed why stationery wooden block of mass 8 kg then they both moves of together in the same straight line calculate the total Momentum just before the impact and just after the impact also calculate the velocity of the combine object
Answers
Mass of object m1 = kg Velocity of object before collision, u1 = 10 m s-1 Mass of wooden block, m2 = 5 kg Velocity of wooden block before collision, u2 = 0 (i) Total momentum before the impact m1u1 + m2u2 = 1 x + 5 x 0 = 10 kg m s-1 According to the law of conservation of momentum (as no net external force acts on the system) : Total momentum after the impact = Total momentum before the impact = 10 kg m s-1 (ii) Mass of combined object, M = mass of object + mass of block = 1 + 5 = 6 kg Let, V = Velocity of the combined object after collision Momentum of combined object = MV = (6V) kg m s-1 Now, momentum of combined object = Total momentum after the impact i.e. 6 V = 10 or
Answer: The velocity of the combined object is 1 m/s. Total Momentum before the impact = Total Momentum after the impact = 10 kg m/s
Explanation:
Mass of an object (m1) = 2 kg
Mass of the wooden block (m2) = 8 kg
Initial velocity of the bullet (u1) = 5 m/s
Initial velocity of the wooden block (u2) = 0 m/s
The object sticks to the wooden block.
∴ Mass of the wooden object with the wooden block (m) = 2 + 8 kg
= 10 kg
Final velocity of the combined object (the object with the wooden block) (v) = ?
According to the Law of Conservation of Momentum,
or, m1u1 + m2u2 = mv
or, 2 X 5 + 8 X 0 = 10 X v
or, 10 = 10 X v
or, v = 10 / 10
or, v = 1
∴ The velocity of the combined object (v) = 1 m/s
STEP 1:
∴ Total Momentum before the impact = m1u1 + m2u2
= 2 X 5 + 8 X 0 kg m/s
= 10 kg m/s
∴Total Momentum after the impact = mv
= 10 X1 kg m/s
= 10 kg m/s
STEP 2:
Total Momentum before the impact = Total Momentum after the impact
= mv
= 10 X1 kg m/s
= 10 kg m/s