Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
Answer:
Check the explanation
Explanation:
This is the step by step explanation to the above question:
![v_i = v [ f_L *(v - v_b) - f_s*(v + v_b)] / [f_L * (v - v_b) + f_s*(v +v_b)]](https://tex.z-dn.net/?f=v_i%20%3D%20v%20%5B%20f_L%20%2A%28v%20-%20v_b%29%20-%20f_s%2A%28v%20%2B%20v_b%29%5D%20%2F%20%5Bf_L%20%2A%20%28v%20-%20v_b%29%20%2B%20f_s%2A%28v%20%2Bv_b%29%5D)
= v * (83.1 * (v-4.3) - 80.7 ( v+4.3))/ [83.1 *(v - 4.3) + 80.7*(v + 4.3)]
v = 344 m/s
vi = 344 * ( 83.1* (344-4.3) - 80.7*(344+4.3) ) / (83.1 *(344 - 4.3) + 80.7*(344 + 4.3))
= 0.74 m/s
Answer:
option D
Explanation:
given.
horizontal velocity of arrow and a ball given as 50 m/s and 44 m/s respectively from the top of a building over flat ground.
In vertical direction, they are both identical
In vertical direction the initial velocity of arrow and a ball is 0 m/s
Their acceleration due to gravity is same for both arrow and a ball 9.8 m/s²
they will react bottom at the same time
time of flight is same for both
now,
In horizontal direction,
distance = speed × time
Since speed is more for arrow, it will travel more horizontal distance at the same time.
the correct answer is option D