Question

A block of mass m=7.6kg, moving on a frictionless surface with speed vi= 7.6m/s, makes a perfectly elastic collision witha block of mass M at rest. After the collision, the 7.6 kg block recoils with a speed of vf=2.5m/s. In the figure, the blocks are in contact for 0.20 s. The average force on the 7.6 kg block, while the two blocks are in contact, is the closest to: 241N, 194N, 289N, 336N, 384N

Answer 1

Keep it simple ..?

F x t = change of momentum (?mv)

For 7.6kg block ..
?(mv) = 7.60kg x (7.6m/s + 2.5m/s) = 76.76 kg.m/s .. (accounting for the change in direction)

F x t = 76.76 kg.m/s
F = 76.76 / 0.20s .. .. .. ?F = 384 N .. (Ans. A)

Answer 2

force on 7.6kg block = change in momentum of 7.6 kb block / time = 7.6 * ( 7.6 + 2.5) / 0.2

= 383.8 N

Answer 3

Answer 4

change in mometum = 7.6( 2.5+7.6) = 76.676 = force *0.2

force = 383.8 N ~ 384 N

Answer 5

Lets say the 7.6 kg block has momentum is right direction.

Hence its initial momentum, m_i = 7.6 * 7.6

It recoils after collision. so it has momentum finally in left direction.

Hence its final momentum, m_f = -(7.6 * 2.5)

Now total change in momentum (impulse) = m_i - m_f

  = 7.6 * 10.1 = 76.76

We know, Impulse = Average Force * Time

Hence, Average Force = Impulse/Time

= 76.76/0.20

Therefore, Average Force = 383.8 N

Which is close to 384 Newton.